Giải phương trình sau a.x^3-2x^2-x+2=0 b.2x(x-5-3x+15=0 c,(x-2)(x+3)=(x-2)(2x+5) Giúp mình với

Đáp án + Giải thích các bước giải:

 a) $x^3 - 2x^2 - x + 2 =0 $

$\Leftrightarrow x^2(x - 2) - (x - 2) = 0$

$\Leftrightarrow (x^2 - 1)(x - 2) = 0$

$\Leftrightarrow (x + 1)(x - 1)(x - 2) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}x+1=0\\x-1=0\\x-2=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=-1\\x=1\\x=2\end{array} \right.\) 

Vậy $S =$`{-1; 1; 2}`

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b) $2x(x - 5 - 3x + 15) = 0$
$\Leftrightarrow 2x(-2x + 10) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}2x=0\\-2x + 10=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=0\\-2x=-10\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\) 

Vậy $S = $`{0; 5}`

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c) $(x - 2)(x + 3) = (x - 2)(2x + 5)$
$\Leftrightarrow (x - 2)(x + 3) - (x - 2)(2x + 5) = 0$

$\Leftrightarrow (x - 2)(x + 3 - 2x - 5) = 0$

$\Leftrightarrow (x - 2)(-x - 2) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}x-2=0\\-x-2=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=2\\-x=2\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 

Vậy $S = $`{-2; 2}`