GIẢI PT Sau: a.25^(x^2-x)+5^(x^2-x+1)-6=0 b.9^(x^2+x-1)-10.3^(x^2+x-2)+1=0 c.3^(2x+5)=3^(x+2)+2

$\begin{array}{l} a)\quad 25^{\displaystyle{x^2 -x}}+5^{\displaystyle{x^2 -x+1}} – 6 = 0\\ \Leftrightarrow 5^{\displaystyle{2(x^2 -x)}}+5.5^{\displaystyle{x^2 -x}} – 6 = 0\\ Đặt\,\,t = 5^{\displaystyle{x^2 -x}}\qquad (t >0)\\ \text{Phương trình trở thành:}\\ \quad t^2 + 5t – 6 -0\\ \Leftrightarrow \left[\begin{array}{l}t = 1\quad (nhận)\\t = -6\quad (loại)\end{array}\right.\\ \text{Ta được:}\\ \quad 5^{\displaystyle{x^2 -x}} = 1\\ \Leftrightarrow x^2 – x = 0\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x = 1\end{array}\right.\\ Vậy\,\,S = \{0;1\}\\ b)\quad 9^{\displaystyle{x^2 +x -1}} -10.3^{\displaystyle{x^2 +x -2}} +1= 0\\ \Leftrightarrow 3.9^{\displaystyle{x^2 +x -1}} – 10.3^{\displaystyle{x^2 +x -1}} +3 =0\\ Đặt\,\,t =3^{\displaystyle{x^2 +x -1}} \qquad (t >0)\\ \text{Phương trình trở thành:}\\ \quad 3t^2 – 10t + 3 =0\\ \Leftrightarrow \left[\begin{array}{l}t = 3\\t = \dfrac13\end{array}\right.\quad (nhận)\\ +)\quad t = 3\\ \Leftrightarrow 3^{\displaystyle{x^2 +x -1}} = 3\\ \Leftrightarrow x^2 + x – 1 = 1\\ \Leftrightarrow x^2 + x -2 =0\\ \Leftrightarrow \left[\begin{array}{l}x = 1\\x = -2\end{array}\right.\\ +)\quad t = \dfrac13\\ \Leftrightarrow 3^{\displaystyle{x^2 +x -1}} = 3^{-1}\\ \Leftrightarrow x^2 + x – 1 = -1\\ \Leftrightarrow x^2 + x = 0\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x = -1\end{array}\right.\\ Vậy\,\,S= \{-2;-1;0;1\}\\ c)\quad 3^{\displaystyle{2x +5}}= 3^{\displaystyle{x +2}} +2\\ \Leftrightarrow 3.3^{\displaystyle{2(x +2)}} – 3^{\displaystyle{x +2}} – 2 =0\\ Đặt\,\,t = 3^{\displaystyle{x +2}} \quad (t >0)\\ \text{Phương trình trở thành:}\\ \quad 3t^2 – t – 2 =0\\ \Leftrightarrow \left[\begin{array}{l}t = 1\quad (nhận)\\t = -\dfrac23\quad (loại)\end{array}\right.\\ \text{Ta được:}\\ \quad 3^{\displaystyle{x +2}} = 1\\ \Leftrightarrow x + 2 =0\\ \Leftrightarrow x = -2\\ Vậy\,\,S =\{-2\} \end{array}$