Bài 1) rút gọn a)(2x-1) ²-2(4+x)(x-4)+(x-4)(x ²+2x+4) b)(1-3x) ²+2(3x-1)(5-3x)+(5+3x) ² Bài 2)tìm x a)x ³-3x ²-x+3=0 b)3x ²-4x-4=0

Đáp án:

 $1)
a) x^3+17\\
b) 16\\
2)
a) {\left[\begin{aligned}x=3\\x=1\\ x=-1\end{aligned}\right.}\\
b)
{\left[\begin{aligned}x=2\\x=\frac{-2}{3}\end{aligned}\right.}\\$

Giải thích các bước giải:

 $1)
a) (2x-1)^2-2(4+x)(x-4)+(x-4)(x^2+2x+4)\\
=4x^2-4x+1-2(x^2-16)+x^3+2x^2+4x-4x^2-8x-16\\
=4x^2-4x+1-2x^2+32+x^3+2x^2+4x-4x^2-8x-16\\
=(4x^2-2x^2+2x^2-4x^2)+(-4x+4x-8x)+(1+32-16)+x^3\\
=x^3+17\\
b) (1-3x)^2+2(3x-1)(5-3x)+(5-3x)^2\\
=(3x-1)^2+2(3x-1)(5-3x)+(5-3x)^2\\
=(3x-1+5-3x)^2\\
=4^2=16$
$2)
a) x^3-3x^2-x+3=0\\
\Leftrightarrow x^2(x-3)-(x-3)=0\\
\Leftrightarrow (x-3)(x^2-1)=0\\
\Leftrightarrow (x-3)(x-1)(x+1)=0\\
\Leftrightarrow {\left[\begin{aligned}x-3=0\\x-1=0\\ x+1=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=3\\x=1\\ x=-1\end{aligned}\right.}$
b)
$3x^2-4x-4=0\\
\Leftrightarrow 3x^2+2x-6x-4=0\\
\Leftrightarrow x(3x+2)-2(3x+2)=0\\
\Leftrightarrow (x-2)(3x+2)=0\\
\Leftrightarrow {\left[\begin{aligned}x-2=0\\3x+2=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=2\\x=\frac{-2}{3}\end{aligned}\right.}\\$