Bài 1 : Rút gọn Bài 2 : Tìm x, biết

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Đáp án+Giải thích các bước giải:

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Bài `1:`

`a)`

`(3x-2)^2-(2x-1)(5x+2)`

`=9x^2-12x+4-10x^2-4x+5x+2`

`=-x^2-11x+6`

`b)`

`(x^2-3)(x^2+3)-5x^2(x+1)^2`

`=x^4-9-5x^2(x^2+2x+1)`

`=x^4-9-5x^4-10x^3-5x^2`

`=-4x^4-10x^3-5x^2-9`

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Bài `2:`

`a)`

`2x^3-18x=0`

`<=>2x(x^2-9)=0`

`<=>2x(x-3)(x+3)=0`

`<=>`$\left[ \begin{array}{l}x=0\\x-3=0\\ x+3=0\end{array} \right.$

`<=>`$\left[ \begin{array}{l}x=0\\x=3\\ x=-3\end{array} \right.$

Vậy `x_1=0;x_2=3;x_3=-3`

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`b)`

`x^2(x-2)-x+2=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0`

`<=>(x-2)(x-1)(x+1)=0`

`<=>`$\left[ \begin{array}{l}x-2=0\\x-1=0\\ x+1=0\end{array} \right.$

`<=>`$\left[ \begin{array}{l}x=2\\x=1\\ x=-1\end{array} \right.$

Vậy `x_1=2;x_2=1;x_3=-1`