Question

Giúp mình với:

Giải phương trình:

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

Answer 1

ĐK:\(-1\le x\le1\)

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

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

\(\Leftrightarrow\frac{2x\sqrt{1-x^2}}{2}=\frac{8x^4-8x^2+1}{2}\)

\(\Leftrightarrow2x\sqrt{1-x^2}=8x^4-8x^2+1\)

\(\Leftrightarrow4x^2\left(1-x^2\right)=64x^8-128x^6+80x^4-16x^2+1\)

\(\Leftrightarrow-\left(2x^2-1\right)^2\left(16x^4-16x^2+1\right)=0\)

Suy ra \(2x^2-1=0\)  hoặc \(16x^4-16x^2+1=0\)

Suy ra \(x=-\frac{1}{\sqrt{2}}\) hoặc \(16\left(x^2-\frac{1}{2}\right)^2-3=0\Rightarrow x=\frac{\sqrt{12}-2}{\sqrt{32}}\) (thỏa)

Answer 2

ĐK:\(-1\le x\le1\)

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

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

\(\Leftrightarrow\frac{2x\sqrt{1-x^2}}{2}=\frac{8x^4-8x^2+1}{2}\)

\(\Leftrightarrow2x\sqrt{1-x^2}=8x^4-8x^2+1\)

\(\Leftrightarrow4x^2\left(1-x^2\right)=64x^8-128x^6+80x^4-16x^2+1\)

\(\Leftrightarrow-\left(2x^2-1\right)^2\left(16x^4-16x^2+1\right)=0\)

Suy ra \(2x^2-1=0\)  hoặc \(16x^4-16x^2+1=0\)

Suy ra \(x=-\frac{1}{\sqrt{2}}\) hoặc \(16\left(x^2-\frac{1}{2}\right)^2-3=0\Rightarrow x=\frac{\sqrt{12}-2}{\sqrt{32}}\) (thỏa)