Ta có \(\sqrt{\left(y-1\right)\left(x-3\right)}\le\frac{x-1+3-y}{2}=1+\frac{x}{2}-\frac{y}{2}\)

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

Nên \(2=\sqrt{\left(x-1\right)\left(3-y\right)}+\sqrt{\left(y-1\right)\left(3-x\right)}\le1+\frac{x}{2}-\frac{y}{2}+1-\frac{x}{2}+\frac{y}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=3-y\\y-1=3-x\end{cases}\Leftrightarrow x+y=4}\)

\(\Rightarrow x^2+y^2-4x-4y+7=0\Leftrightarrow\left(x+y\right)^2-2xy-4\left(x+y\right)+7=0\)

\(\Leftrightarrow xy=\frac{7}{2}\)

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\(ĐK:x\ge1\)

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

\(\Leftrightarrow\left(2x-4\right)+\left(\sqrt{x+2}-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow2\left(x-2\right)+\frac{-3\left(x-2\right)}{\sqrt{x+2}+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(2+\frac{-3}{\sqrt{x+2}+2\sqrt{x-1}}\right)=0\)

\(TH1:x-2=0\Leftrightarrow x=2\)

\(TH2:2=\frac{3}{\sqrt{x+2}+2\sqrt{x-1}}\)(*)

Mà ta có: \(\sqrt{x+2}+2\sqrt{x-1}\ge\sqrt{1+2}+0=\sqrt{3}\)

\(\Rightarrow\frac{3}{\sqrt{x+2}+2\sqrt{x-1}}\le\sqrt{3}< 2\)

Như vậy, (*) vô nghiệm

Vậy phương trình có một nghiệm duy nhất là 2.

Chọn câu A