Question

giải phương trình: \(\sqrt{3x+1}\)+2\(\sqrt[3]{19x+8}\)=2x²+x+5

Answer 1

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(2x^2-2x+\left(x+1-\sqrt{3x+1}\right)+2\left(x+2-\sqrt[3]{19x+8}\right)=0\)

\(\Leftrightarrow2x^2-2x+\dfrac{x^2-x}{x+1+\sqrt[]{3x+1}}+\dfrac{\left(x+7\right)\left(x^2-x\right)}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{19x+8}+\sqrt[3]{\left(19x+8\right)^2}}=0\)

\(\Leftrightarrow\left(x^2-x\right)\left(2+\dfrac{1}{x+1+\sqrt[]{3x+1}}+\dfrac{x+7}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{19x+8}+\sqrt[3]{\left(19x+8\right)^2}}\right)=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)