√(x ²+x-1) + √(x-x ²+1) = x ²-x+1 giải phường trình( giải kỹ một tý nha) mấy bạn không giải thì đừng phá nhá

$\begin{array}{l} \sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} = {x^2} - x + 2\\ DK:\,\,\,\left\{ \begin{array}{l} {x^2} + x - 1 \ge 0\\ x - {x^2} + 1 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x \ge \frac{{ - 1 + \sqrt 5 }}{2}\\ x \le \frac{{ - 1 - \sqrt 5 }}{2} \end{array} \right.\\ \frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2} \end{array} \right. \Leftrightarrow \frac{{\sqrt 5 - 1}}{2} \le x < \frac{{\sqrt 5 + 1}}{2}\\ \Leftrightarrow \sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} = {x^2} - x + 2\\ Xet\,\,\,VT = \sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} \\ Ap\,\,\,dung\,\,bdt\,\,BunhiaCopxki\,\,ta\,\,duoc:\\ {\left( {\sqrt {{x^2} + x - 1} .1 + \sqrt {x - {x^2} + 1} .1} \right)^2} \le \left[ {{{\left( {\sqrt {{x^2} + x - 1} } \right)}^2} + {{\left( {\sqrt {x - {x^2} + 1} } \right)}^2}} \right]\left( {{1^2} + {1^2}} \right)\\ \Leftrightarrow {\left( {\sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} } \right)^2} \le 2\left( {{x^2} + x - 1 + x - {x^2} + 1} \right)\\ \Leftrightarrow {\left( {\sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} } \right)^2} \le 4x.\\ \Leftrightarrow \sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} \le 2\sqrt x \,\,\,\,\left( {x > 0\,\,tmdkxd} \right)\\ Ap\,\,dung\,\,bdt\,\,Co - si\,\,ta\,\,co:\,\,\,x + 1 \ge 2\sqrt x \\ \Rightarrow \sqrt {{x^2} + x - 1} + \sqrt {x - {x^2} + 1} \le 2\sqrt x \Leftrightarrow x + 1\\ Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow x = 1.\\ Xet\,\,\,VP = {x^2} - x + 2 = {x^2} - 2x + 1 + x + 1 = {\left( {x - 1} \right)^2} + \left( {x + 1} \right) \ge x + 1\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1.\\ \Rightarrow VP = VT \Leftrightarrow x = 1.\\ Vay\,\,x = 1\,\,\,\,la\,\,\,\,nghiem\,\,\,cua\,\,pt. \end{array}$