Bài 2)tìm x a)5x.(x-1)+x.(x+17)=0 b)x.(5x^2+6)-x.(4x^2+6)=0 c)3x.(x-3)^2-3x.(x+3)^2=0 d)7-9x+2x^2=0 e)4x^3-4x=0

Đáp án:

a/ $x=0$ $\text{hoặc}$ $x=-2$

b/ $x=0$

c/ $x=0$

d/ $x=\dfrac{7}{2}$ $\text{hoặc}$ $x=1$

e/ $x=0$ $\text{hoặc}$ $x=±1$

Giải thích các bước giải:

a/ $5x(x-1)+x(x+17)=0$

$⇔ 5x^2-5x+x^2+17x=0$

$⇔ 6x^2+12x=0$

$⇔ 6x(x+2)=0$

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

b/ $x(5x^2+6)-x(4x^2+6)=0$

$⇔ 5x^3+6x-4x^3-6x=0$

$⇔ x^3=0$

$⇔ x=0$

c/ $3x(x-3)^2-3x(x+3)^2=0$

$⇔ 3x(x^2-6x+9)-3x(x^2+6x+9)=0$

$⇔ 3x^3-18x^2+27x-3x^3-18x^2-27x=0$

$⇔ -36x^2=0$

$⇔ x=0$

d/ $7-9x+2x^2=0$

$⇔ 2x^2-7x-2x+7=0$

$⇔ x(2x-7)-(2x-7)=0$

$⇔ (2x-7)(x-1)=0$

$⇔ \left[ \begin{array}{l}x=\dfrac{7}{2}\\x=1\end{array} \right.$

e/ $4x^3-4x=0$

$⇔ 4x(x^2-1)=0$

$⇔ 4x(x-1)(x+1)=0$

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