Question

$\sqrt{\sqrt{2}-x}=x\sqrt{\frac{3}{2}(\sqrt{2}+x)}$

Answer 1

ĐKXĐ: \(-\sqrt{2}\le x\le\sqrt{2}\)

- Với \(x< 0\Rightarrow\) vế trái ko âm, vế phải âm, pt vô nghiệm

- Với \(x\ge0\) bình phương 2 vế ta được:

\(\sqrt{2}-x=\dfrac{3}{2}x^2\left(\sqrt{2}+x\right)\)

\(\Leftrightarrow3x^3+3\sqrt{2}x^2+2x-2\sqrt{2}=0\)

\(\Leftrightarrow27x^3+27\sqrt{2}x^2+18x-18\sqrt{2}=0\)

\(\Leftrightarrow\left(3x\right)^3+3.\left(3x\right)^2.\sqrt{2}+3.\left(\sqrt{2}\right)^2.3x+\left(\sqrt{2}\right)^3-20\sqrt{2}=0\)

\(\Leftrightarrow\left(3x+\sqrt{2}\right)^3=20\sqrt{2}\)

\(\Leftrightarrow3x+\sqrt{2}=\sqrt[3]{20\sqrt{2}}\)

\(\Leftrightarrow x=\dfrac{\sqrt[3]{20\sqrt{2}}-\sqrt{2}}{3}\)