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

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1. 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)
   $3x^2-4x-4=0\\
   \Leftrightarrow 3x^2+2x-6x-4=0\\
   \Leftrightarrow x(3x+2)-2(3x+2)=0\\
   \Leftrightarrow (x-2)(3x+2)=0\\
   \Leftrightarrow {\left[\begin{aligned}x-2=0\\3x+2=0\end{aligned}\right.}\\
   \Leftrightarrow {\left[\begin{aligned}x=2\\x=\frac{-2}{3}\end{aligned}\right.}\\$

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