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Bài 1:
a) 5(x-3)-4=2(x-1)
\(\Leftrightarrow5x-15-4=2x-2\)
\(\Leftrightarrow5x-19-2x+2=0\)
\(\Leftrightarrow3x-17=0\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)
Vậy: \(x=\frac{17}{3}\)
b) 5-(6-x)=4(3-2x)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow-1+x-12+8x=0\)
\(\Leftrightarrow-13+9x=0\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\frac{13}{9}\)
Vậy: \(x=\frac{13}{9}\)
c) (3x+5)(2x+1)=(6x-2)(x-3)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)
\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy:x=1
Bài 2:
a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)
\(\Leftrightarrow4x-10x-15x-3x+60=0\)
\(\Leftrightarrow-24x+60=0\)
\(\Leftrightarrow-24x=-60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy: \(x=\frac{5}{2}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)
\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)
\(\Leftrightarrow-3x=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow3x=-48\)
\(\Leftrightarrow x=-16\)
Vậy: x=-16
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)
\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)
\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)
\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow-13x=-143\)
\(\Leftrightarrow x=11\)
Vậy: x=11
e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)
\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)
\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)
\(\Leftrightarrow45x-18-24-28x+60x-420=0\)
\(\Leftrightarrow77x-462=0\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
Vậy:x=6
Bài 3:
a) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)
Vì \(2\ne0\)
nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)
b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)
c) \(\left(2x+1\right)\left(x^2+2\right)=0\)
Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)
Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta lại có \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)
Ta có: \(4\ne0\)(4)
Từ (3) và (4) suy ra
2x-1=0
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
Bài 4:
a) \(\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)
\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)
\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)
\(\Leftrightarrow-8x^2+40x-32=0\)
\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)
\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)
Vì \(-8\ne0\)
nên \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{1;4\right\}\)
e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)
\(\Leftrightarrow7x^2+58x+115=0\)
\(\Leftrightarrow7x^2+23x+35x+115=0\)
\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)
\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)
Bài 5:
a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)
b) \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)
\(\Leftrightarrow3x^2-3=0\)
\(\Leftrightarrow3\left(x^2-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-1\right\}\)
c) \(x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)
Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)
Từ (5) và (6) suy ra
\(\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy: x=-1
a) \(\left(x+2\right)^2-9=0\)
\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)
\(=>\left(x-1\right).\left(x+5\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x= 1 hoặc x= -5
b) \(x^2-2x+1=25\)
\(=>x^2-2.x.x+1^2=25\)
\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)
\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)
\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy x= 6 hoặc x= -4
c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)
\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)
\(=>4x\left(x-1\right)-4x^2+25-1=0\)
\(=>4x\left(x-1\right)-4x^2+24=0\)
\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)
..................... tắc ròi -.-"
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)
\(=>x^3+27-x^3-3x=15\)
\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)
Vì \(3>0=>4-x=0=>x=4\)
Vậy x= 4
e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)
\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)
\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)
\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)
\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'
a/ \(\left(x+2\right)^2-9=0\)
<=> \(\left(x+2-3\right)\left(x+2+3\right)=0\)
<=> \(\left(x-1\right)\left(x+5\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
b/ \(x^2-2x+1=25\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Bài 2 :
a. A = 2 ( x3 + y3 ) - 3 ( x2 + y2 ) với x + y = 1
=> A = 2 ( x + y ) ( x2 - xy + y2 ) - 3 [ ( x + y )2 - 2xy ]
=> A = 2 [ ( x + y )2 - 3xy ] - 3 ( 1 - 2xy )
=> A = 2 ( 1 - 3xy ) - 3 + 6xy
=> A = 2 - 6xy - 3 + 6xy
=> A = - 1
B = x3 + y3 + 3xy với x + y = 1
=> B = ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
=> B = ( x + y )3 - 3xy ( x + y - 1 )
=> B = 13 - 3xy . 0
=> B = 1
Bài 1.
a) ( x - 1 )3 + ( 2 - x )( 4 + 2x + x2 ) + 3x( x + 2 ) = 16
<=> x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x = 16
<=> 9x + 7 = 16
<=> 9x = 9
<=> x = 1
b) ( x + 2 )( x2 - 2x + 4 ) - x( x2 - 2 ) = 15
<=> x3 + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 7
<=> x = 7/2
c) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 15
<=> ( x - 3 )[ ( x - 3 )2 - ( x2 + 3x + 9 ) + 9( x2 + 2x + 1 ) = 15
<=> ( x - 3 )( x2 - 6x + 9 - x2 - 3x - 9 ) + 9x2 + 18x + 9 = 15
<=> ( x - 3 ).(-9x) + 9x2 + 18x + 9 = 15
<=> -9x2 + 27x + 9x2 + 18x + 9 = 15
<=> 45x + 9 = 15
<=> 45x = 6
<=> x = 6/45 = 2/15
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 = 11
<=> x = -11/25
Bài 2.
a) A = 2( x3 + y3 ) - 3( x2 + y2 )
= 2( x + y )( x2 - xy + y2 ) - 3x2 - 3y2
= 2( x2 - xy + y2 ) - 3x2 - 3y2
= 2x2 - 2xy + 2y2 - 3x2 - 3y2
= -x2 - 2xy - y2
= -( x2 + 2xy + y2 )
= -( x + y )2
= -(1)2 = -1
b) B = x3 + y3 + 3xy
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 + 3xy
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 3x2y + 3xy2 - 3xy )
= ( x + y )3 - 3xy( x + y - 1 )
= 13 - 3xy( 1 - 1 )
= 1 - 3xy.0
= 1
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
a) \(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)
b) \(3x\left(x-2\right)-5x+10=0\)
\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
c) \(4x\left(x+3\right)-x^2+9=0\)
\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)
f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)
g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)
a) 4x3 - 9x = 0
<=> x( 4x2 - 9 ) = 0
<=> x( 2x - 3 )( 2x + 3 ) = 0
<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0
<=> x = 0 hoặc x = ±3/2
b) 3x( x - 2 ) - 5x + 10 = 0
<=> 3x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( 3x - 5 ) = 0
<=> x - 2 = 0 hoặc 3x - 5 = 0
<=> x = 2 hoặc x = 5/3
c) 4x( x + 3 ) - x2 + 9 = 0
<=> 4x( x + 3 ) - ( x2 - 9 ) = 0
<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0
<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0
<=> ( x + 3 )( 4x - x + 3 ) = 0
<=> ( x + 3 )( 3x + 3 ) = 0
<=> x + 3 = 0 hoặc 3x + 3 = 0
<=> x = -3 hoặc x= -1
d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )
<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0
<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0
<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0
<=> ( x - 4 ).3x = 0
<=> x - 4 = 0 hoặc 3x = 0
<=> x = 4 hoặc x = 0
e) 16x2 - 25 = ( 4x - 5 )( 2x + 1 )
<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0
<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0
<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0
<=> ( 4x - 5 )( 2x + 4 ) = 0
<=> 4x - 5 = 0 hoặc 2x + 4 = 0
<=> x = 5/4 hoặc x = -2
f) ( x + 1/5 )2 = 64/9
<=> ( x + 1/5 )2 = ( ±8/3 )2
<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3
<=> x = 37/15 hoặc x = -43/15
g) 9( x + 2 )2 = ( x + 3 )2
<=> 32( x + 2 )2 - ( x + 3 )2 = 0
<=> [ 3( x + 2 ) ]2 - ( x + 3 )2 = 0
<=> ( 3x + 6 )2 - ( x + 3 )2 = 0
<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0
<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0
<=> ( 2x + 3 )( 4x + 9 ) = 0
<=> 2x + 3 = 0 hoặc 4x + 9 = 0
<=> x = -3/2 hoặc x = -9/4
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`