\(^2\)-1-x(2x+1)=0

b, x\(^2\)-7x+12=0...">

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5 tháng 10 2021

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)

19 tháng 3 2020

a, (3x-1)2 - (x+3)2 = 0

<=> [(3x-1)-(x+3)][(3x-1)+(x+3)] = 0

<=> (3x-1-x-3)(3x-1+x+3) = 0

<=> (2x-4)(4x+2) = 0

=> 2x-4=0 hoặc 4x+2=0

=> 2x =4 hoặc 4x = -2

=> x = 2 hoặc x = \(\frac{-1}{2}\)

19 tháng 3 2020

\(\begin{array}{l} a){\left( {3x - 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow \left( {3x - 1 + x + 3} \right)\left[ {3x - 1 - x - 3} \right] = 0\\ \Leftrightarrow \left( {4x + 2} \right)\left( {2x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 2 = 0\\ 2x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{2}\\ x = 2 \end{array} \right.\\ b){x^3} - \dfrac{x}{{49}} = 0\\ \Leftrightarrow 49{x^3} - x = 0\\ \Leftrightarrow x\left( {49{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 49{x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \pm \dfrac{1}{7} \end{array} \right.\\ c){x^2} - 7x + 12 = 0\\ \Leftrightarrow {x^2} - 3x - 4x + 12 = 0\\ \Leftrightarrow x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 3 = 0\\ x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 4 \end{array} \right.\\ d)4{x^2} - 3x - 1 = 0\\ \Leftrightarrow 4{x^2} + x - 4x - 1 = 0\\ \Leftrightarrow x\left( {4x + 1} \right) - \left( {4x + 1} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 1 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{4}\\ x = 1 \end{array} \right.\\ e){x^3} - 2x - 4 = 0\\ \Leftrightarrow {x^3} - 4x + 2x - 4 = 0\\ \Leftrightarrow x\left( {{x^2} - 4} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left[ {x\left( {x + 2} \right) + 2} \right] = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ {x^2} + 2x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ {x^2} + 2x + 2x = 0\left( {VN} \right) \end{array} \right.\\ f){x^3} + 8{x^2} + 17x + 10 = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 7x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 5x + 2x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left[ {x\left( {x + 5} \right) + 2\left( {x + 5} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ x + 5 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 5\\ x = - 2 \end{array} \right. \end{array}\)

1 tháng 10 2020

a) (2x - 3)2 = (x + 5)2

=> 4x2 - 12x + 9 = x2 + 10x + 25

=> 4x2 - 12x + 9 - (x2 + 10x + 25) = 0

=> 3x2 - 22x - 16 = 0

=> 3x2 - 24x + 2x - 16 = 0

=> 3x(x - 8) + 2(x - 8) = 0

=> (3x + 2)(x - 8) = 0

=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)

b) x2(x - 1) - 4x2 + 8x - 4 = 0

=> x2(x - 1) - (2x  - 2)2 = 0

=> x2(x - 1) - [2(x- 1)]2 = 0

=> x2(x - 1) - 4(x - 1)2 = 0

=> (x - 1)(x2 - 4(x - 1) = 0

=> (x - 1)(x2 - 4x + 4) = 0

=> (x - 1)(x - 2)2 = 0

=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

c) x2 + 7x + 12 = 0

=> x2 + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 4)(x + 3) = 0

=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)

d) x2 + 3x - 18 = 0

=> x2 + 6x - 3x - 18 = 0

=> x(x + 6) - 3(x + 6) = 0

=> (x - 3)(x + 6) = 0

=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

e) x(x + 6) - 7x - 42 = 0

=> x(x + 6) - 7(x + 6) = 0

=> (x - 7)(x + 6) = 0

=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)

1 tháng 10 2020

1. ( 2x - 3 )2 = ( x + 5 )2

<=> ( 2x - 3 )2 - ( x + 5 )2 = 0

<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0

<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0

<=> ( x - 8 )( 3x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

2. x2( x - 1 ) - 4x2 + 8x - 4 = 0

<=> x2( x - 1 ) - ( 4x2 - 8x + 4 ) = 0

<=> x2( x - 1 ) - 4( x2 - 2x + 1 ) = 0

<=> x2( x - 1 ) - 4( x - 1 )2 = 0

<=> ( x - 1 )[ x2 - 4( x - 1 ) ] = 0

<=> ( x - 1 )( x2 - 4x + 4 ) = 0

<=> ( x - 1 )( x - 2 )2 = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

3. x2 + 7x + 12 = 0

<=> x2 + 3x + 4x + 12 = 0

<=> x( x + 3 ) + 4( x + 3 ) = 0

<=> ( x + 3 )( x + 4 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

4. x2 + 3x - 18 = 0

<=> x2 - 3x + 6x - 18 = 0

<=> x( x - 3 ) + 6( x - 3 ) = 0

<=> ( x - 3 )( x + 6 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

5. x( x + 6 ) - 7x - 42 = 0

<=> x( x + 6 ) - 7( x + 6 ) = 0

<=> ( x + 6 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)

4 tháng 3 2020

(x2 + x  + 1)(6 - 2x) = 0

<=> 6 - 2x = 0 (do x2 + x + 1 > 0)

<=> 2x = 6

<=> x = 3

Vậy S = {3}

(8x - 4)(x2 + 2x + 2) = 0

<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)

<=> 8x = 4

<=> x = 1/2 

Vậy S  = {1/2}

x3 - 7x + 6 = 0

<=> x3 - x - 6x + 6 = 0

<=> x(x2 - 1) - 6(x - 1) = 0

<=> x(x - 1)(x + 1) - 6(x - 1) = 0

<=> (x2 + x - 6)(x - 1) = 0

<=> (x2 + 3x - 2x - 6)(x - 1) = 0

<=> (x + 3)(x - 2)(x - 1) = 0

<=> x + 3 = 0

hoặc x - 2 = 0

hoặc x  - 1 = 0

<=> x = -3

hoặc x = 2

hoặc x = 1

Vậy S = {-3; 1; 2}

x5 - 5x3 + 4x = 0

<=> x(x4 - 5x2 + 4) = 0

<=> x(x4 - x2 - 4x2 + 4) = 0

<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0

<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0

<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x  + 1 = 0

<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1

Vậy S = {-2; -1; 0; 1; 2}

4 tháng 3 2020

+ Ta có: \(\left(x^2+x+1\right).\left(6-2x\right)=0\)

 - Ta lại có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

- Vì \(x^2+x+1>0\forall x\)mà \(\left(x^2+x+1\right).\left(6-2x\right)=0\)

  \(\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)\)

Vậy \(S=\left\{3\right\}\)

+ Ta có: \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)

 - Ta lại có: \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\)

 - Vì \(x^2+2x+2>0\forall x\)mà \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)

   \(\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)

+ Ta có: \(x^3-7x+6=0\)

       \(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0\)

       \(\Leftrightarrow x^2.\left(x-1\right)+x.\left(x-1\right)-6.\left(x-1\right)=0\)

       \(\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0\)

       \(\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0\) 

       \(\Leftrightarrow\left(x-1\right).\left[x.\left(x-2\right)+3.\left(x-2\right)\right]=0\)

       \(\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0\)

       \(\Leftrightarrow x=1\left(TM\right)\)hoặc \(x=2\left(TM\right)\)hoặc \(x=-3\left(TM\right)\)

 Vậy \(S=\left\{-3;1;2\right\}\)

 + Ta có: \(x^5-5x^3+4x=0\)

        \(\Leftrightarrow x.\left(x^4-5x^2+4\right)=0\)

       \(\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0\)

       \(\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0\)

       \(\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0\)

       \(\Leftrightarrow x=0\left(TM\right)\)

hoặc  \(x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)\)

hoặc \(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)\)

Vậy \(S=\left\{-2;-1;0;1;2\right\}\)

!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!      

26 tháng 7 2017

a)\(x^2-2x-24=0\Leftrightarrow x^2-2x+1-25=0\)

\(\Leftrightarrow\left(x-1\right)^2-5^2=0\Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)

b)\(x^2+8x+12=0\Leftrightarrow x^2+8x+16-4=0\)

\(\Leftrightarrow\left(x+4\right)^2-2^2=0\Leftrightarrow\left(x+4-2\right)\left(x+4+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\Leftrightarrow\hept{\begin{cases}x=-2\\x=-6\end{cases}}\)

c)\(4x^2+4x-63=0\Leftrightarrow4x^2+4x+1-64=0\)

\(\Leftrightarrow\left(2x+1\right)^2-8^2=0\Leftrightarrow\left(2x+1-8\right)\left(2x+1+8=0\right)\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+9\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)

4 tháng 4 2018

a.Ta có : \(\dfrac{x^2-4x+4}{x^3-2x^2-4x+8}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\)

Để \(\dfrac{1}{x+2}>0\) thì 1 và x+2 cùng dấu

mà 1>0

=>x + 2 > 0 <=> x > 2

\(\Rightarrow S=\left\{x|x>2\right\}\)

b, Ta có : \(x^2\ge0\Rightarrow x^2+1>0\)

Để \(\dfrac{7-8x}{x^2+1}>0\) thì 7 - 8x và \(x^2+1\) cùng dấu

\(x^2+1>0\Rightarrow7-8x>0\Leftrightarrow x< \dfrac{7}{8}\)

\(\Rightarrow S=\left\{x|x< \dfrac{7}{8}\right\}\)

c. Ta có bảng xét dấu:

x -\(\infty\) -1 -\(\dfrac{1}{2}\) +\(\infty\)
x+1 - 0 + +
2x+1 - - 0 +
\(\dfrac{2x+1}{x+1}\) + \(//\) - 0 +

4 tháng 4 2018

Bổ xung câu c:

Vậy : \(-1< x\le\dfrac{-1}{2}\)

27 tháng 8 2017

\(a,\)\(x^4-4x^3+4x^2=0\)

\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)

\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)

10 tháng 2 2020

\(a.2x^2+7x-9=0\\ \Leftrightarrow2\left(x^2+\frac{7}{2}x-\frac{9}{2}\right)=0\\\Leftrightarrow x^2+\frac{7}{2}x-\frac{9}{2}=0\\ \Leftrightarrow x^2+\frac{9}{2}x-x-\frac{9}{2}=0\\\Leftrightarrow x\left(x+\frac{9}{2}\right)-\left(x+\frac{9}{2}\right)=0\\\Leftrightarrow \left(x-1\right)\left(x+\frac{9}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+\frac{9}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{9}{2}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{1;-\frac{9}{2}\right\}\)

10 tháng 2 2020

\(b.x^2-4x+3=0\\\Leftrightarrow x^2-x-3x+3=0\\ \Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\\Rightarrow \left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{1;3\right\}\)

7 tháng 11 2021

e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)

\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)

=> 2 TH

*3x+7=0               *10x-4=0

=>3x=-7               =>10x=4

=>x=-7/3              =>x=4/10=2/5

vậy x=-7/3 hoặc x=2/5

g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)

\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)

\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)

\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)

\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)

=> 2 TH

*-(x+3)=0          *3x-5=0

=>-x=-3            =>3x=5  

=x=3                =>x=5/3

h)\(x^2-x^2+x-1=0\)

\(\Rightarrow0+x-1=0\)

\(\Rightarrow x-1=0\)

=>x=0+1

=>x=1

vậy x=1

k, x(x+ 16) - 7x - 42 = 0

=>x^2+16x-7x-42=0

=>x^2+9x-42=0

vì x^2>0

do đó x^2+9x-42>0

nên o có gt nào của x t/m y/cầu đề bài

m)x^2+7x+12=0

=>x^2+3x++4x+12=0

=>x(x+3)+4(x+3)=0

=>(x+4).(x+3)=0

=>2 TH

=> *x+4=0

=>x=-4

vậy x=-4

*x+3=0

=>x=-3

vậy x=-3

n)x^2-7x+12=0

=>x^2-4x-3x+12=0

=>x(x-4)-3(x-4)=0

=>(x-3).(x-4)=0

=>2 TH

*x-3=0=>x=0+3=>x=3

*x-4=0=>x=0+4=>x=4

vậy x=3 hoặc x=4

7 tháng 11 2021

a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1

b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\) \(\Leftrightarrow x^2+5x-2x-10-5=1x+5\) \(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0 \(\Leftrightarrow\) \(x^2+2x-20=0\) \(\Leftrightarrow x^2+2x-10x-20=0\) \(\Leftrightarrow\) (x\(^2\) + 2x) - (10x +...
Đọc tiếp

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

\(\Leftrightarrow\) \(x^2+2x-20=0\)

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4

\(\Leftrightarrow\) 4x - 4 = 0

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

0
12 tháng 8 2019

b) \(7x\left(x-2\right)-\left(x-2\right)=0\) 

<=>  \(\left(7x-1\right)\left(x-2\right)=0\)

=> x=1/7  hoặc x=2

c) <=>  (2x-1)3   =0 

=> x=1/2

d)<=>  \(\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)

<=>  \(\left(2x-3\right)\left(x+3\right)=0\)

=> x=3/2  hoặc x=-3

e) <=>\(x^2\left(x+5\right)+9\left(x+5\right)=0\)

<=> \(\left(x+5\right)\left(x^2+9\right)=0\)

=> x=-5

f) \(x^3-6x^2-x+30=0\)

<=>\(x^3+2x^2-8x^2-16x+15x+30=0\)

<=>\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

<=>\(\left(x+2\right)\left(x^2-5x-3x+15\right)=0\)

<=> \(\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)

=> x=-2 hoặc x=5 hoặc x=3