a/ ĐKXĐ: \(x\ge\sqrt[3]{2}\)
\(\Leftrightarrow\sqrt{x^3-2}-\left(2x-1\right)+x-1-\sqrt[3]{x^2-1}=0\)
\(\Leftrightarrow\frac{x^3-2-\left(2x-1\right)^2}{\sqrt{x^3-2}+2x-1}+\frac{\left(x-1\right)^3-\left(x^2-1\right)}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\frac{x^3-4x^2+4x-3}{\sqrt{x^3-2}+2x-1}+\frac{x^3-4x^2+3x}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2-x+1\right)}{\sqrt{x^3-2}+2x-1}+\frac{\left(x-3\right)\left(x^2-x\right)}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2-x+1}{\sqrt{x^3-2}+2x-1}+\frac{x^2-x}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}\right)=0\)
\(\Rightarrow x=3\)
b/ Đặt \(\sqrt[3]{35-x^3}=a\)
\(\Rightarrow\left\{{}\begin{matrix}ax\left(a+x\right)=30\\x^3+a^3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ax\left(a+x\right)=90\\x^3+a^3=35\end{matrix}\right.\)
\(\Rightarrow x^3+a^3+3ax\left(a+x\right)=125\)
\(\Leftrightarrow\left(x+a\right)^3=125\)
\(\Leftrightarrow x+a=5\)
\(\Leftrightarrow a=5-x\)
\(\Leftrightarrow\sqrt[3]{35-x^3}=5-x\)
\(\Leftrightarrow35-x^3=125-75x+15x^2-x^3\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow...\)
