[tex]4^{x^2+3x}-34.2^{x^2+5x+1}+16.4^{2x+2}=0[/tex]
$\iff 2^{2(x^2+3x)}+64 \cdot 4^{2x+1}=34 \cdot 2^{x^2+5x+1}$ (1)
Đặt $\left\{\begin{matrix} u=2^{x^2+3x} \\ v=2^{2x+1} \end{matrix} \right.$
$\implies u.v=2^{x^2+5x+1}$
$(1) \iff u^2+64v^2=34u.v$
$\iff u^2-32uv+64v^2-2uv=0$
$\iff u(u-32v)-2v(u-32v)=0$
$\iff (u-32v)(u-2v)=0$
$\iff \left[\begin{matrix} u=32v \\ u=2v \end{matrix} \right.$
$\iff \left[\begin{matrix} 2^{x^2+3x}=32 \cdot 2^{2x+1} \\ 2^{x^2+3x} =2\cdot 2^{2x+1} \end{matrix} \right.$
$\iff \left[\begin{matrix} 2^{x^2+3x}=2^5 \cdot 2^{2x+1} \\ 2^{x^2+3x} =2\cdot 2^{2x+1} \end{matrix} \right.$
$\iff \left[\begin{matrix} 2^{x^2+3x}= 2^{2x+6} \\ 2^{x^2+3x} = 2^{2x+2} \end{matrix} \right.$
$\iff \left[\begin{matrix} x^2+x-6=0 \\ x^2+x-2=0 \end{matrix} \right.$
Tới đây em giải tiếp nha, em tham khảo thêm
Chinh phục kì thi THPTQG môn toán 2022 nha.