Đáp án:
a, x = $\frac{-11}{2}$
b, x = {$\frac{7}{3}; \frac{8}{3}; 2$}
Giải thích các bước giải:
a, $(\frac{4}{5})^{2x+7}$ = $\frac{625}{256}$
⇔ $(\frac{4}{5})^{2x+7}$ = $(\frac{5}{4})^4$
⇔ $(\frac{4}{5})^{2x+7}$ = $(\frac{4}{5})^{-4}$
⇔ 2x + 7 = -4 ⇔ 2x = -11 ⇔ x = $\frac{-11}{2}$
b, $(3x-7)^{2005} = (3x-7)^{2003}$
⇔ $(3x-7)^{2003}.[(3x-7)^{2}-1]$ = 0
⇔ \(\left[ \begin{array}{l}(3x-7)^{2003}=0\\(3x-7)^{2}-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x-7=0\\\left[ \begin{array}{l}3x-7=1\\3x-7=-1\end{array} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{7}{3}\\\left[ \begin{array}{l}x=\frac{8}{3}\\x=2\end{array} \right.\end{array} \right.\)
