Giải phương trình $9^{x^2+x-1}-{10.3}^{x^2+x-2}+1=0$ .

 $9^{x^2+x-1}-{10.3}^{x^2+x-2}+1=0$

$\iff {9.9}^{x^2+x-2}-{10.3}^{x^2+x-2}+1=0$(*)

Đặt  $t=3^{x^2+x-2}\left(t>0\right)$ thì phương trình (*) trở thành:

$(*)\iff 9t^2-10t+1=0$

Û (9t − 1)(t − 1) = 0

$\iff \left[\begin{array}{l}9t-1=0\\ t-1=0\end{array}\right.\iff \left[\begin{array}{l}t=\frac{1}{9}\\ t=1\end{array}\right.$

• Với   $t=\frac{1}{9}\Rightarrow 3^{x^2+x-2}=\frac{1}{9}$

$\Rightarrow x^2+x-2=-2$$$\begin{gathered} \iff x\left(x+1\right)=0 \\ \iff \left[\begin{array}{l}x=0\\ x+1=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=0\\ x=-1\end{array}\right. \end{gathered}$$

• Với  $t=1\Rightarrow 3^{x^2+x-2}=1$

$\Rightarrow x^2+x-2=0$.

$$\begin{gathered} \iff \left(x+2\right)\left(x-1\right)=0 \\ \iff \left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=-2\\ x=1\end{array}\right. \end{gathered}$$