Giải:
Đặt \(y=b\sqrt{1-x}\)
Ta có: \(\sqrt{a+y}=1+\sqrt{a-y}\)
\(\Leftrightarrow\sqrt{a+y}-\sqrt{a-y}=1\)
\(\Leftrightarrow\left(\sqrt{a+y}-\sqrt{a-y}\right)^2=1\)
\(\Leftrightarrow a+y-2\cdot\sqrt{a+y}\cdot\sqrt{a-y}+a-y=1\)
\(\Leftrightarrow2a-2\sqrt{a^2-y^2}=1\)
\(\Leftrightarrow2\sqrt{a^2-y^2}=2a-1\)
\(\Leftrightarrow\sqrt{a^2-y^2}=\dfrac{2\left(a-\dfrac{1}{2}\right)}{2}=a-\dfrac{1}{2}\)
\(\Leftrightarrow a^2-y^2=\left(a-\dfrac{1}{2}\right)^2=a^2-a+\dfrac{1}{4}\)
\(\Leftrightarrow y^2=a-\dfrac{1}{4}\)
\(\Leftrightarrow y=\sqrt{a-\dfrac{1}{4}}\)
\(\Leftrightarrow b\sqrt{1-x}=\sqrt{a-\dfrac{1}{4}}\)
\(\Leftrightarrow\sqrt{1-x}=\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\)
\(\Leftrightarrow1-x=\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)
\(\Leftrightarrow x=1-\left(\dfrac{\sqrt{a-\dfrac{1}{4}}}{b}\right)^2\)
Vậy....................
