a) \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)

\(pt\Leftrightarrow\sqrt{-x^2+2x+1+1}+\sqrt{-x^2-6x-9+1}=1+\sqrt{3}\)

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

Dễ thấy: \(VT\le2< 1+\sqrt{3}=VP\) (vô nghiệm)

b)\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)

\(pt\Leftrightarrow\sqrt{9x^2-6x+1+1}+\sqrt{45x^2-30x+5+4}=\sqrt{-9x^2+6x-1+9}\)

\(\Leftrightarrow\sqrt{\left(3x-1\right)^2+1}+\sqrt{5\left(3x-1\right)^2+4}=\sqrt{-\left(3x-1\right)^2+9}\)

Dễ thấy: \(VT\ge1+\sqrt{4}=3=VP\)

Đẳng thức xảy ra khi \(x=\dfrac{1}{3}\)

đề sai sao ý, cái căn thứ 2

Ta có :

\(\sqrt{9x^2-6x+2}=\sqrt{\left(9x^2-6x+1\right)+1}=\sqrt{\left(3x-1\right)^2+1}\ge\sqrt{1}=1\)

\(\sqrt{45x^2-30x+9}=\sqrt{5\left(9x^2-6x+1\right)+4}=\sqrt{5\left(3x-1\right)^2+4}\ge\sqrt{4}=2\)

\(\sqrt{6x-9x^2+8}=\sqrt{-\left(9x^2-6x+1\right)+9}=\sqrt{-\left(3x-1\right)^2+9}\le3\)

\(\Rightarrow VT\ge3\ge VP\)

mÀ đề lại cho \(VT=VP\) \(\Rightarrow\hept{\begin{cases}\sqrt{\left(3x-1\right)^2+1}=1\\\sqrt{\left(3x-1\right)^2+4}=2\\\sqrt{-\left(3x-1\right)^2+9}=3\end{cases}\Rightarrow x=\frac{1}{3}}\)

Vậy \(x=\frac{1}{3}\)

x=1/3 nha

X=1/3 đấy !!!!