Question

Giúp mk giải pt câu này với ạ:

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

Answer 1

\(\sqrt[3]{x+2}=\sqrt{x-2}\)  \(\left(DK:x\ge2\right)\)

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

\(\left(x+2\right)^2=\left(x-2\right)^3\)

\(x^2+4x+4=x^3-6x^2+12x-8\)

\(x^3-7x^2+8x-12=0\)

\(\left(x-6\right)\left(x^2-x+2\right)=0\)

=> x=6 (tm)

Answer 2

ĐKXĐ: \(x\ge2\)

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

\(\Rightarrow\left(\sqrt[3]{x+2}\right)^6=\left(\sqrt[]{x-2}\right)^6\)

\(\Leftrightarrow\left(x+2\right)^2=\left(x-2\right)^3\)

\(\Leftrightarrow x^3-7x^2+8x-12=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x+2\right)=0\)

\(\Rightarrow x=6\)