Question

Giải phương trình : x^3+1=2.³\(\sqrt[]{}\)2x-1

Answer 1

ĐK: \(x\ge\dfrac{1}{2}\)

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

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

\(\Leftrightarrow\left(x-1\right)\left[x^2+x+1-4.\dfrac{1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\right]=0\)

Dễ thấy \(x^2+x+1-\dfrac{4}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\ne0\)

\(\Rightarrow x=1\)