Помогите решить плиз

ОДЗ: x+1≥0
x≥ -1

$2^{ \sqrt{x+1} }-1\ \textless \ \frac{3*2^2}{2^{\sqrt{x+1}}} \\ \\ y=2^{\sqrt{x+1}} \\ \\ y-1\ \textless \ \frac{12}{y} \\ \\ y-1- \frac{12}{y}\ \textless \ 0$

ОДЗ: y≠0
$\frac{y^2-y-12}{y}\ \textless \ 0$
Разложим на множители:
y²-y-12=0
D=1+48=49
y₁=(1-7)/2=-3
y₂=(1+7)/2=4

y²-y-12=(y+3)(y-4)

y(y+3)(y-4)<0<br>y=0 y=-3 y=4
- + - +
------- -3 ---------- 0 ----------- 4 -----------
\\\\\\\\\ \\\\\\\\\\\\\
y=5 + + + | +
y=1 + + - | -
y= -1 - + - | +
y= -4 - - - | -

{y< -3
{ 0
1) y < -3
$2^{ \sqrt{x+1} }\ \textless \ -3$
нет решений.

2) 0 $0\ \textless \ 2^{ \sqrt{x+1} }\ \textless \ 4 \\ a) 2^{ \sqrt{x+1} }\ \textgreater \ 0 \\ x \geq -1 \\ \\ b)2^{ \sqrt{x+1} }\ \textless \ 4 \\ 2^{ \sqrt{x+1} }\ \textless \ 2^2 \\ \sqrt{x+1}\ \textless \ 2 \\ x+1\ \textless \ 2^2 \\ x+1\ \textless \ 4 \\ x\ \textless \ 4-1 \\ x\ \textless \ 3$

{x≥ -1
{x<3<br> \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------- -1 ------------ 3 ----------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[-1; 3)
Ответ: [-1; 3).