Question

[Nguyễn.T.Huệ]

giải phương trình : căn bậc 4(1-x)+căn bậc 4(x+15)=2

Answer 1

[Nguyen]

\(\sqrt[4]{1-x}+\sqrt[4]{x+15}=2\)(ĐK: \(-15\le x\le1\))

\(\Leftrightarrow\dfrac{\sqrt{1-x^2}}{\sqrt[4]{1+x}}+\dfrac{\sqrt{x+15}-4}{\sqrt[4]{x+15}+2}=0\)

\(\Leftrightarrow\dfrac{1-x^2}{\sqrt{1+x}}=\dfrac{x-1}{\left(\sqrt[4]{x+15}+2\right)^2}\)

\(\Leftrightarrow\left(x-1\right)\left[\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{\left(1+x\right)}{\sqrt{1-x}}\right]=0\)

\(\Leftrightarrow x=1\)(Vì \(\dfrac{1}{\left(\sqrt[4]{x+15}+2\right)^2}-\dfrac{1+x}{\sqrt{1-x}}\)\(\ne0\)

Vậy pt có nghiệm là x=1.