Question

[Dreya Gray]

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

Answer 1

[Mr Lazy]

ĐK: \(-1\le x\le1\)

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

Pt đã cho thành: \(x^3+y^3=xy\sqrt{2}\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-xy\sqrt{2}=0\)

Đặt \(x+y=a;\text{ }xy=b\)

Ta có: \(a^2-2b=1;\text{ }a^3-3ab-b\sqrt{2}=0\)

\(a^2-2b=1\Rightarrow b=\frac{a^2-1}{2}\)

\(a^3-3ab-b\sqrt{2}=0\Leftrightarrow a^3-3a.\frac{a^2-1}{2}-\frac{a^2-1}{2}\sqrt{2}=0\)

\(\Leftrightarrow-a^3-\sqrt{2}a^2+3a+\sqrt{2}=0\)

\(\Leftrightarrow\left(a-\sqrt{2}\right)\left(a+\sqrt{2}-1\right)\left(a+\sqrt{2}+1\right)=0\)

\(\Leftrightarrow a=\sqrt{2}\text{ hoặc }a=1-\sqrt{2}\text{ hoặc }a=-1-\sqrt{2}\)

\(+a=\sqrt{2};\text{ }b=\frac{2-1}{2}=\frac{1}{2}\)\(\Rightarrow x+y=\sqrt{2};\text{ }xy=\frac{1}{2}\)

=> x,y là 2 nghiệm của pt \(X^2-\sqrt{2}X+\frac{1}{2}=0\Leftrightarrow\left(X-\frac{1}{\sqrt{2}}\right)^2=0\Leftrightarrow X=\frac{1}{\sqrt{2}}\)

\(\Rightarrow x=\sqrt{1-x^2}=\frac{1}{\sqrt{2}}\text{ (nhận)}\)

\(+a=1-\sqrt{2}\Rightarrow b=1-\sqrt{2}\Rightarrow x+y=xy=1-\sqrt{2}\)

=> x, y là 2 nghiệm của pt \(X^2-\left(1-\sqrt{2}\right)X+1-\sqrt{2}=0\)

\(\Leftrightarrow X=\frac{1-\sqrt{2}+\sqrt{2\sqrt{2}-1}}{2}>0\text{ hoặc }X=\frac{1-\sqrt{2}-\sqrt{2\sqrt{2}-1}}{2}