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ND
8 tháng 2 2018
a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
26 tháng 8 2018
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
SM
26 tháng 8 2018
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
16 tháng 8 2017
ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A
16 tháng 8 2017
a) \(\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2=4\)
\(\Rightarrow\left(x-3\right)^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)
\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{5;-1\right\}\)
b) \(x^2-2x=24\)
\(\Rightarrow x.\left(x+2\right)=24\)
\(\Rightarrow x.\left(x+2\right)=4.6\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
NC
19 tháng 7 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x\right)^2-3^2=0\)
\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
Vậy.........
c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x=-255\)
\(\Leftrightarrow x=-127,5\)
Vậy.............
có j sai xót mong m.n bỏ qua☺
19 tháng 7 2018
a) \(25x^2-9=0\)
<=> \(\left(5x\right)^2=9\)
<=> \(\left(5x\right)^2=3^2\)
<=> \(5x=3\)
<=> \(x=\frac{3}{5}\)
b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)
<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)
<=> \(x^2+8x+16-x^2+1=16\)
<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)
<=> \(8x+17=16\)
<=> \(8x=-1\)
<=> \(x=\frac{-1}{8}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)
<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)
<=> \(2x+245=0\)
<=> \(2x=-245\)
<=> \(x=\frac{-245}{2}\)
S
13 tháng 10 2018
\(a.x^4-16x^2=0\Leftrightarrow\left(x^2+4x\right)\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
\(b.\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
NL
13 tháng 10 2018
a) x4 - 16x2 = 0
<=> x2 ( x2 - 16 ) = 0
<=> \(\left[{}\begin{matrix}x^2=0\\x^2-16=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
Vậy...
b) ( x - 5)3 - x + 5 = 0
<=> ( x - 5)3 - (x - 5) = 0
<=> (x - 5) [ (x - 5)2 - 1] =0
<=> \(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
c) 5(x - 2) = x2 - 4
<=> 5(x - 2) - (x2 - 4) = 0
<=> (x - 2)( 5 - x - 2) = 0
<=> (x - 2)( 3 - x ) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy...
d) x - 3 = (3 - x)2
<=> x - 3 - (x - 3)2 = 0
<=> (x - 3)(1 - x + 3) = 0
<=> (x - 3)( 4 - x ) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy...
e) x2 (x - 5) + 5 - x = 0
<=> x2 (x - 5) - (x - 5) = 0
<=> (x2 - 1)( x - 5) = 0
<=> \(\left[{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
,
10 tháng 2 2020
\(\left(x-5\right)\left(x-1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-5-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy............
10 tháng 2 2020
\(5\left(x+3\right)\left(x-2\right)-3\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow5\left(x^2+x-6\right)-3\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow2x^2-16x-60=0\)
\(\Leftrightarrow x^2-8x-30=0\)
làm tiếp nhé!!!!!
XO
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
<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)
\(x\left(x-5\right)+3\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm pt là: \(S=\left\{5;-3\right\}\)
x(x-5)+3(x-5)=0
=>(x-5)(x+3)=0
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)