a) \(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)
Đặt \(\sqrt{x^2-3x+3}=a;\sqrt{x^2-3x+6}=b\left(a;b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\b^2-a^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(b+a\right)\left(b-a\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+a=3\\b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\) (nhận)
\(\Rightarrow\sqrt{x^2-3x+3}=1\)
\(\Leftrightarrow x^2-3x+3=1\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\) (nhận)
b) \(\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)
Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\\left(b^2+2b+1\right)+b^2-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2\left(b-1\right)\left(b+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) (vì \(b+2>0\)) (nhận)
\(\Rightarrow\sqrt{2+x-x^2}=1\)
\(\Leftrightarrow2+x-x^2=1\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)
