Question

giải pt \(\sqrt[3]{x+6}+\sqrt{x-1}=x^2-1\)

Answer 1

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt[3]{x+6}-2+\sqrt{x-1}-1=x^2-4\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt[]{x-1}+1}=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}=x+2\left(1\right)\end{matrix}\right.\)

Xét (1), do \(x\ge1\Rightarrow\left\{{}\begin{matrix}x+2\ge3\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}< \dfrac{1}{4}+\dfrac{1}{1}< 3\\\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)