Question

[Minh]

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

Answer 1

[Cô Hoàng Huyền]

ĐK: \(-1\le x\le1\)

Đặt \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}}\left(a,b\ge0\right)\)

\(\Rightarrow\hept{\begin{cases}a^2-b^2=2x\\a^2+b^2=2\end{cases}}\)

Từ đó ta có pt: \(\left(a-1\right)\left(b+1\right)=a^2-b^2\)

\(\Leftrightarrow ab+a-b-1=a^2-b^2\)

\(\Leftrightarrow2ab+2a-2b-2=2a^2-2b^2\)

\(\Leftrightarrow2ab+2a-2b-a^2-b^2=2a^2-2b^2\)

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

\(\Leftrightarrow\left(a-b\right)^2-2\left(a-b\right)+2\left(a-b\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-b-2+2a+2b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(3a+b-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\3a+b-2=0\end{cases}}\)

Với a = b ta có: \(\sqrt{1+x}=\sqrt{1-x}\Rightarrow x=0\)

Với 3a + b - 2 = 0 ta có: \(3\sqrt{1+x}+\sqrt{1-x}-2=0\)

\(\Leftrightarrow3\sqrt{1+x}=2-\sqrt{1-x}\)

\(\Leftrightarrow9\left(1+x\right)=5-x-4\sqrt{1-x}\)

\(\Leftrightarrow-10x-4=4\sqrt{1-x}\)

\(\Rightarrow100x^2+80x+16=16-16x\)

\(\Rightarrow100x^2+96x=0\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{96}{100}\end{cases}}\)  (tm)

Vậy pt có nghiệm \(x=0\) hoặc \(x=-\frac{96}{100}.\)