tìm x biết a. x(x-1)=0 b. $3x^{2}$ -6x=0 c. x(x-6)+10(x-6)=0 d. $x^{3}$ -x=0 e.3x(x-10)=x-10 f.x(x+7)= 4x+48

Các bước giải:

`a. x.(x -1) = 0`

$⇔ \left[ \begin{array}{l}x=0\\x -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=0\\x=1\end{array} \right.$

Vậy `S = {0; 1}`

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`b. 3x² -6x = 0`

`⇔ 3x.(x -2) = 0`

$⇔ \left[ \begin{array}{l}3x=0\\x -2=0\end{array} \right. \left[ \begin{array}{l}x=0\\x=1\end{array} \right. $

`Vậy S = {0; 2}`

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`c. x.(x -6) +10.(x -6) = 0`

`⇔ (x -6).(x +10) = 0`

$⇔ \left[ \begin{array}{l}x -6=0\\x +10=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=6\\x=-10\end{array} \right.$

Vậy `S = {6; -10}`

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`d. x³ -x = 0`

`⇔ x.(x² -1) = 0`

$⇔ \left[ \begin{array}{l}x=0\\x² -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=0\\x=±1\end{array} \right.$

Vậy S = {0; ±1}`

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`e. 3x.(x -10) = x -10`

`⇔ 3x.(x -10) -(x -10) = 0`

`⇔ (x -10).(3x -1) = 0`

$⇔ \left[ \begin{array}{l}x -10=0\\3x -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=10\\x=\dfrac{1}{3}\end{array} \right.$

Vậy `S = {10; 1/3}`

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`f. x.(x +7)= 4x +48`

`⇔ x² +7x -4x -48 = 0`

`⇔ x² +3x -48 = 0`

`⇔ (x +3/2)² -201/4 = 0`

`⇔ (x +3/2 -(\sqrt{201})/2).(x +3/2 +(\sqrt{201})/2) = 0`

$⇔ \left[ \begin{array}{l}x +3/2 -(\sqrt{201})/2=0\\x +3/2 +(\sqrt{201})/2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-3 +\sqrt{201}}{2}\\x=\dfrac{-3 -\sqrt{201}}{2}\end{array} \right.$

Vậy `S = {`$\dfrac{-3 +\sqrt{201}}{2}; \dfrac{-3 -\sqrt{201}}{2}$`}`

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