Question

[đàm quang vinh]

giải phương trình

\(x^4+2x^3+2x^2-2x+1=\left(x^3+x\right)\)\(\sqrt{\frac{1-x^2}{x}}\)

Answer 1

[Lê Tài Bảo Châu]

\(x^4+2x^3+2x^2-2x+1=\left(x^3+x\right)\sqrt{\frac{1-x^2}{x}}\)ĐK: \(0< x< 1\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2-2x+1\right)=x\left(x^2+1\right)\sqrt{\frac{\left(1-x\right)\left(1+x\right)}{x}}\)

\(\Leftrightarrow x^2\left(x+1\right)^2+\left(1-x\right)^2=\left(x^2+1\right)\sqrt{x\left(x+1\right)\left(1-x\right)}\left(1\right)\)

Đặt  \(\hept{\begin{cases}x\left(x+1\right)=a>0\\1-x=b>0\end{cases}}\)\(\Rightarrow x^2+1=a+b\)

\(\Rightarrow\left(1\right)\Leftrightarrow a^2+b^2=\left(a+b\right)\sqrt{ab}\)

\(\Leftrightarrow a^2+b^2-a\sqrt{ab}-b\sqrt{ab}=0\)

\(\Leftrightarrow a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+b\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a^3}-\sqrt{b^3}\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\left(a+\sqrt{ab}+b\right)=0\)

Vì \(a+\sqrt{ab}+b=\left(\sqrt{a}+\frac{\sqrt{b}}{2}\right)^2+\frac{3\sqrt{b}}{4}>0;\forall a,b>0\)

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=0\)

\(\Leftrightarrow a=b\)

\(\Rightarrow x^2+x=1-x\)

\(\Leftrightarrow x^2+2x-1=0\)

\(\Delta=8\)

=> pt có 2 nghiệm pb \(^{\orbr{\begin{cases}x=-1+\sqrt{2}\left(tm\right)\\x=-1-\sqrt{2}\left(loai\right)\end{cases}}}\)

Vậy ...