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\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\)
\(2x^2-x+\frac{1}{3}-\frac{2}{3}x=0\)
\(2x^2-\frac{5}{3}x+\frac{1}{3}=0\)
\(6x^2-5x+1=0\)
\(6x^2-3x-2x+1\)
\(3x\left(2x-1\right)-\left(2x-1\right)=0\)
\(\left(3x-1\right)\left(2x-1\right)=0\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)

a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }


a) x(x-1) - (x+1)(x+2) = 0
x\(^2\)- x -x\(^{^2}\)-2x +x+2=0
-2x+2=0
-2x=0+2
-2x=2
x=-1
Vậy x bằng -1

\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

(x2-1)3-(x4+x2+1)(x2-1)=0
<=> (x2-1)[(x2-1)2-x4-x2-1]=0
<=> (x-1)(x+1)[x4-2x2+1-x4-x2-1]=0
<=> (x-1)(x+1)(-3x2)=o
<=> 3x2(x-1)(x+1)=0
=> x1=0; x2=-1; x3=1
Đáp số: x1=0; x2=-1; x3=1
Ta có: \(\left(x-3\right)\left(x-1\right)-x\left(2-x\right)=0\)
=>\(x^2-4x+3-2x+x^2=0\)
=>\(2x^2-6x+3=0\)
=>\(x^2-3x+\dfrac{3}{2}=0\)
=>\(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}+\dfrac{3}{2}=0\)
=>\(\left(x-\dfrac{3}{2}\right)^2-\dfrac{3}{4}=0\)
=>\(\left(x-\dfrac{3}{2}\right)^2=\dfrac{3}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{3}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{3}}{2}\\x=\dfrac{3-\sqrt{3}}{2}\end{matrix}\right.\)