Đáp án:
a.$x=\pm4$
b.$x=2$
Giải thích các bước giải:
a.Ta có:
$3a^3(x^2-1)^4:3a^3(x^2-1)^3=15$
$\to (3a^3:3a^3)\cdot ((x^2-1)^4:(x^2-1)^3)=15$
$\to x^2-1=15$
$\to x^2=16$
$\to x=\pm4$
b.Ta có:
$x^3(2x-1)^{m+2}:x^3(2x-1)^{m-1}-3^5:3^2=0$
$\to (x^3:x^3)\cdot ((2x-1)^{m+2}:(2x-1)^{m-1})-3^{5-2}=0$
$\to (2x-1)^{(m+2)-(m-1)}-3^3=0$
$\to (2x-1)^3=3^3$
$\to 2x-1=3$
$\to 2x=4$
$\to x=2$
