ĐK: \(1\le x\le4\)
Đặt \(\sqrt{4-x}+\sqrt{2x-2}=a>0\)
\(\Rightarrow4-x+2x-2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=a^2\)
\(\Leftrightarrow x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=a^2-2\)
Pt đã cho trở thành:
\(5+a^2-2=4a\Leftrightarrow a^2-4a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\)
TH1: \(a=1\Rightarrow\sqrt{4-x}+\sqrt{2x-2}=1\)
\(\Leftrightarrow x+2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=1\)
\(\Leftrightarrow x+1+2\sqrt{\left(4-x\right)\left(2x-2\right)}=0\)
Do \(x\ge1\Rightarrow VT>0\Rightarrow\) vô nghiệm
TH2: \(a=3\Rightarrow\sqrt{4-x}+\sqrt{2x-2}=3\)
Áp dụng BĐT Bunhiacốpxki ta có:
\(\sqrt{4-x}+\sqrt{2x-2}=\sqrt{4-x}+\sqrt{2}\sqrt{x-1}\le\sqrt{\left(1+2\right)\left(4-x+x-1\right)}=3\)
Dấu "=" xảy ra khi và chỉ khi: \(\sqrt{4-x}=\dfrac{\sqrt{x-1}}{\sqrt{2}}\) \(\Leftrightarrow8-2x=x-1\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
