Question

[Phan Mai Hoa]

GIẢI PHƯƠNG TRÌNH BẰNG CÁCH ĐẶT ẨN PHỤ \(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)

Answer 1

[Huyền]

Đặt \(t=\sqrt{x}+\sqrt{1-x}\)\(\Rightarrow t^2=1+2\sqrt{x\left(1-x\right)}\)(\(t\ge0\))

\(pt:1+\frac{2}{3}\sqrt{x\left(1-x\right)}=\sqrt{x}+\sqrt{1-x}\)(\(0\le x\le1\))

\(\Leftrightarrow\frac{1}{3}\left(1+2\sqrt{x\left(1-x\right)}\right)+\frac{2}{3}=\sqrt{x}+\sqrt{1-x}\)

\(\Leftrightarrow\frac{1}{3}t^2+\frac{2}{3}=t\)

\(\Leftrightarrow t^2+2-3t=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}1=\sqrt{x}+\sqrt{1-x}\\2=\sqrt{x}+\sqrt{1-x}\end{matrix}\right.\)

TH1:\(1=\sqrt{x}+\sqrt{1-x}\Leftrightarrow1=1+\sqrt{x\left(1-x\right)}\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

TH2:\(2=\sqrt{x}+\sqrt{1-x}\Leftrightarrow4=1+\sqrt{x\left(1-x\right)}\Leftrightarrow3=\sqrt{x\left(1-x\right)}\)

\(-x^2+x-9=0\)(vô nghiệm)

Vậy pt có nghiệm x = 0 , x = 1 .