Giải các phương trình sau:

a) ${32}^{\frac{x+5}{x-7}}=0,25\cdot {128}^{\frac{x+17}{x-3}}$  ;

a) Điều kiện: $\left\{\begin{array}{l}x-7\ne 0\\ x-3\ne 0\end{array}\right.\iff \left\{\begin{array}{l}x\ne 7\\ x\ne 3\end{array}\right.$

Ta có: ${32}^{\frac{x+5}{x-7}}=0,25\cdot {128}^{\frac{x+17}{x-3}}\iff 2^{5\cdot \frac{x+5}{x-7}}=2^{-2}\cdot 2^{7\cdot \frac{x+17}{x-3}}\iff 2^{\frac{5\left(x+5\right)}{x-7}}=2^{-2+\frac{7\left(x+17\right)}{x-3}}$

$\iff \frac{5\left(x+5\right)}{x-7}=-2+\frac{7\left(x+17\right)}{x-3}$

$\iff \frac{5\left(x+5\right)\left(x-3\right)}{\left(x-7\right)\left(x-3\right)}=\frac{-2\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x-3\right)}+\frac{7\left(x+17\right)\left(x-7\right)}{\left(x-7\right)\left(x-3\right)}$

$$\begin{gathered} \Rightarrow 5\left(x+5\right)\left(x-3\right)=-2\left(x-7\right)\left(x-3\right)+7\left(x+17\right)\left(x-7\right) \\ \iff 5\left(x^2+2x-15\right)=-2\left(x^2-10x+21\right)+7\left(x^2+10x-119\right) \\ \iff 5x^2+10x-75=-2+7x^2+70x-833 \\ \iff 5x^2+10x-75=5x^2+90x-875\iff 80x=800\iff x=10 \end{gathered}$$

(thỏa mãn điều kiện).

Vậy nghiệm của phương trình là x = 10.