a, \(pt\Leftrightarrow\sqrt[3]{x}=\sqrt[3]{x-1}+\sqrt[3]{1-2x}\)
\(\Leftrightarrow x=x-1+1-2x+3\sqrt[3]{\left(x-1\right)\left(1-2x\right)}\left(\sqrt[3]{x-1}+\sqrt[3]{1-2x}\right)\)
\(\Rightarrow x=-x+3\sqrt[3]{\left(x-1\right)\left(1-2x\right)}.\sqrt[3]{x}\)
\(\Leftrightarrow2x=3\sqrt[3]{-2x^3+3x^2-x}\)
\(\Leftrightarrow8x^3=27\left(-2x^3+3x^2-x\right)\)
\(\Leftrightarrow62x^3-81x^2+27x=0\)
\(\Leftrightarrow x.\left(62x^2-81x+27\right)=0\)
Do \(62x^2-81x+27>0\forall x\) nên \(x=0\)
Vậy phương trình có nghiệm \(x=0\)
b, \(\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2x-3}\)
\(\Leftrightarrow x-1+x-2+3\sqrt[3]{\left(x-1\right)\left(x-2\right)}\left(\sqrt[3]{x-1}+\sqrt[3]{x-2}\right)=2x-3\)
\(\Rightarrow3\sqrt[3]{\left(x-1\right)\left(x-2\right)}.\sqrt[3]{2x-3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\frac{3}{2}\end{matrix}\right.\)
Thử lại ta được \(x=1;x=2;x=\frac{3}{2}\) là nghiệm của phương trình
