\(ĐKXĐ:-1\le x\le2\)
\(PT\Leftrightarrow3-x+x^2+2+x-x^2-2\sqrt{\left(3-x+x^2\right)\left(2+x-x^2\right)}=1\\ \left(ĐK:3-x+x^2>2+x-x^2\ge0\right)\)\(\Leftrightarrow\sqrt{\left(3-x+x^2\right)\left(2+x-x^2\right)}=2\\ \Leftrightarrow\left(3-x+x^2\right)\left(2+x-x^2\right)=4\)
Bạn tách hết ra rồi giải PT bậc 4
Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\\left(1+b\right)^2+b^2-5=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2b^2+2b-4=0\)
\(\Leftrightarrow2\left(b-1\right)\left(b+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2+x-x^2}=1\)
\(\Leftrightarrow1=2+x-x^2\)
\(\Leftrightarrow x^2-x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\left(n\right)\)
Vậy phương trình có 2 nghiệm phân biệt . . .
