Question

giải giúp mk bài này nha mn. Arigatou 

Giải phương trình :\(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)

Answer 1

CÁi  này easy mà .-.

\(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)

\(\Leftrightarrow\frac{\frac{\left(7-x\right)-\left(x-5\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)

\(\Leftrightarrow\frac{\frac{-2\left(x-6\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{\frac{-2}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+1\right)=0\)

\(\Rightarrow x-6=0\Rightarrow x=6\)