Giải thích các bước giải:
$\sqrt{5x-1}+\sqrt[3]{9-x}=2x^2+3x-1$
Đặt $\sqrt[3]{9-x}=u\to x=-u^3+9$
$\to \sqrt{5\left(-u^3+9\right)-1}+u=2\left(-u^3+9\right)^2+3\left(-u^3+9\right)-1$
$\to \sqrt{5\left(-u^3+9\right)-1}+u=2\left(-u^3+9\right)^2+3\left(-u^3+9\right)-1$
$\to v+\left(-\sqrt[3]{\frac{v^2-44}{5}}\right)=2\left(-\left(-\sqrt[3]{\frac{v^2-44}{5}}\right)^3+9\right)^2+3\left(-\left(-\sqrt[3]{\frac{v^2-44}{5}}\right)^3+9\right)-1$
$\to v-\sqrt[3]{\frac{v^2-44}{5}}=\frac{2v^4}{25}+\frac{19v^2}{25}-\frac{4708}{25}+188$
$\to v-\sqrt[3]{\frac{v^2-44}{5}}-v=\frac{2v^4}{25}+\frac{19v^2}{25}-\frac{4708}{25}+188-v$
$\to \left(-\sqrt[3]{\frac{v^2-44}{5}}\right)^3=\left(\frac{2v^4}{25}+\frac{19v^2}{25}-\frac{4708}{25}+188-v\right)^3$
$\to v=2,\:v\approx \:-1.17079\dots $
$\to v=2\to \sqrt{5\left(-u^3+9\right)-1}=2$
$\to u=2\to \sqrt[3]{9-x}=2:\to x=1$
