Question

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

Answer 1

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

Dễ thấy \(\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}>0\)

Vậy (1)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy S={1}

Answer 2

\(\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0\)

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

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

\(\Leftrightarrow\sqrt[3]{x-1}=0\Rightarrow x=1\)

(Do \(\dfrac{2\sqrt[3]{\left(x-1\right)^2}}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}+1>0\) \(\forall x\) )