Решить иррациональное уравнение, нужна ваша помощь.Номера ввиде фотографий.

A)
$\sqrt{x^2+5x+4}\ \textgreater \ x+2$

Рассматриваем две системы неравенств:
$1) \left \{ {{x+2\ \textless \ 0} \atop {x^2+5x+4 \geq 0}} \right. \\ \\ 2) \left \{ {{x+2 \geq 0} \atop {x^2+5x+4\ \textgreater \ (x+2)^2}} \right.$

1)
a) x+2<0<br> x< -2

b) x²+5x+4≥0
D=25-16=9
x₁=(-5-3)/2=-4
x₂=(-5+3)/2=-1
\\\\\\\\\\\\\\\\\\\\\\\
-------- -4 ------- -2 ------- -1 -----------
\\\ + \\\\\ \\\\+\\\\\\\\
x∈(-∞; -4]

2)
a) x+2≥0
x≥ -2

b) x²+5x+4>(x+2)²
x²+5x+4>x²+4x+4
x²-x²+5x-4x>4-4
x>0
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
--------- -2 ---------- 0 -----------------
\\\\\\\\\\\\\\\\\\\\
x∈(0; +∞)

В итоге x∈(-∞; -4]U(0; +∞)

b)
$\sqrt{3x^2+5x-2} \leq x+2$

Система неравенств:
{x+2≥0
{3x²+5x-2≤(x+2)²
{3x²+5x-2≥0

a) x+2≥0
x≥ -2

b) 3x²+5x-2 ≤ (x+2)²
3x² +5x-2 ≤ x² +4x+4
3x² - x²+5x-4x-2-4 ≤ 0
2x² +x -6 ≤ 0
D=1+48=49
x₁=(-1-7)/4= -2
x₂=(-1+7)/4=1.5
+ - +
--------- -2 ----------- 1.5 --------------
\\\\\\\\\\\\\\
x∈[-2; 1.5]

c) 3x²+5x-2≥0
D=25+24=49
x₁=(-5-7)/6= -2
x₂=(-5+7)/6=2/6=1/3
+ - +
------- -2 ------------ 1/3 ----------------
\\\\\\\\\ \\\\\\\\\\\\\\\\\
x∈(-∞; -2]U[1/3; +∞)

{x≥ -2
{x∈[-2; 1.5]
{x∈(-∞; -2]U[1/3; +∞)
\\\\\\\\\\\\\\\\\\\\\\\\\\\\
------ -2 --------- 1/3 -------- 1.5 -------------
/////// //////////////////////////////////

x∈[1/3; 1.5]
Ответ: [1/3; 1.5]