Đáp án:
Giải thích các bước giải:
10) $(2x-1)^2+(2-x)(2x-1)=0\\\Leftrightarrow (2x-1)((2x-1)+2-x)\\\Leftrightarrow (2x-1)(x+1)=0\\\Leftrightarrow \left[\begin{array}{l}2x-1=0\Rightarrow x=\cfrac{1}{2}\\x+1=0\Rightarrow x=-1\end{array}\right.$
Vậy, $x=\{\cfrac{1}{2}; -1\}$
11) $x^3+1=x(x+1)\\\Leftrightarrow (x+1)(x^2-x+1)=x(x+1)\\\Leftrightarrow (x+1)(x^2-x+1-x)=0\\\Leftrightarrow (x+1)(x^2-2x+1)=0\\\Leftrightarrow (x+1)(x-2)^2=0\\\Leftrightarrow \left[\begin{array}{l}x=-1\\x=2\end{array}\right.$
Vậy, $x=\{-1; 2\}$
16) $x^3+x^2+x+1=0\\\Leftrightarrow x^3+1=-x^2-x\\\Leftrightarrow (x+1)(x^2-x+1)=-x(x+1)\\\Leftrightarrow (x+1)(x^2-x+1+x+1)=0\\\Leftrightarrow (x+1)(x^2+2)=0\\\Leftrightarrow x=-1\, (Do\, x^2+2=0\, là\, vô\, nghiệm)$
Vậy, $x=-1$
