Đáp án: $x\in\{-1,\dfrac{-1}{4}\}$
Giải thích các bước giải:
$(8x+5)^2(4x+3)(2x+1)=9$
$\rightarrow (64x^2+80x+25)(8x^2+10x+3)=9$
$\rightarrow (8(8x^2+10x)+25)(8x^2+10+3)=9$
Đặt $8x^2+10x=t$
$\rightarrow (8t^2+25)(t+3)=9$
$\rightarrow 8t^3+49t+75=9$
$\rightarrow 8t^2+49t+66=0$
$\rightarrow (8t+33)(t+2)=0$
$\rightarrow t\in\{-2,\dfrac{-33}{8}\}$
$+)t=-2\rightarrow 8x^2+10x=-2$
$\rightarrow 4x^2+5x+1=0$
$\rightarrow (4x+1)(x+1)=0$
$\rightarrow x\in\{-1,\dfrac{-1}{4}\}$
$+)t=-\dfrac{33}{4}\rightarrow 8x^2+10x=-\dfrac{33}{8}$
$\rightarrow 64x^2+80x+33=0$
$\rightarrow (8x+5)^2+8=0\rightarrow$Phương trình vô nghiệm
