Đáp án:
Bài 1: a) $x=\left \{ \dfrac{-1+\sqrt{1+4\sqrt[3]{6}}}{2};\dfrac{-1+\sqrt{1+4\sqrt[3]{6}}}{2} \right \}$
b) $x=\left \{ 0;-\sqrt{2\sqrt{2}-2} \right \}$
Bài 2: $x\in \varnothing $
Giải thích các bước giải:
Bài 1:
$a)(x^{2}+x)^{2}(x^{2}+x)-6=0\\\Leftrightarrow (x^{2}+x)^{3}-6=0\\\Leftrightarrow (x^{2}+x)^{3}=6\\\Leftrightarrow x^{2}+x=\sqrt[3]{6}\\\Leftrightarrow x^{2}+x-\sqrt[3]{6}=0\\\Leftrightarrow x=\dfrac{-1±\sqrt{1^{2}-4.1.(-\sqrt[3]{6})}}{2.1}\\\Leftrightarrow x=\dfrac{-1±\sqrt{1+4\sqrt[3]{6}}}{2}$
Vậy $x=\left \{ \dfrac{-1+\sqrt{1+4\sqrt[3]{6}}}{2};\frac{-1+\sqrt{1+4\sqrt[3]{6}}}{2} \right \}$
$b) x^{3}+2\sqrt{2x^{2}}+2x=0\\\Leftrightarrow x^{3}+2\sqrt{2}.|x|+2x=0\\\Leftrightarrow \left[ \begin{array}{l}x^{3}+2\sqrt{2}x+2x=0,x\geq 0\\x^{3}+2\sqrt{2}.(-x)+2x=0,x<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=0\\x=-\sqrt{2\sqrt{2}-2}\end{array} \right.$
Vậy $x=\left \{ 0;-\sqrt{2\sqrt{2}-2} \right \}$
Bài 2: $x\in \varnothing $
