Question

Giải phương trình :

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

Answer 1

ĐK:\(\frac{1}{2}\leq x\leq \frac{1}{\sqrt{2}}\)

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

Đặt \(\sqrt{2x-1}=a; \sqrt{1-2x^2}=b(a,b\geq 0)\)

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

PT trở thành:
\(a+b=2\sqrt{\frac{a^2+b^2}{2}}\)

\(\Rightarrow (a+b)^2=4.\frac{a^2+b^2}{2}=2(a^2+b^2)\)

\(\Rightarrow a^2+b^2+2ab=2(a^2+b^2)\)

\(\Rightarrow 2ab=a^2+b^2\Rightarrow (a-b)^2=0\Rightarrow a=b\)

\(\Rightarrow a^2=b^2\)

Hay \(2x-1=1-2x^2\)

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

\(\Rightarrow x=\frac{\pm \sqrt{5}-1}{2}\)

Kết hợp với ĐKXĐ suy ra \(x=\frac{\sqrt{5}-1}{2}\)