√(5x+5)=√(5+2x+x²) , 5x+5≥0,5x≥-5,x≥-1
x²+2x+5≥0, (x+1)²-1+5≥0, (x+1)²+4≥0, x∈R
5x+5=5+2x+x²
x²-3x=0, x(x-3)=0, x=0 ∨ x=3
Otvet: 3
3
√(5x+5)=√(5+2x+x²) , 5x+5≥0,5x≥-5,x≥-1
x²+2x+5≥0, (x+1)²-1+5≥0, (x+1)²+4≥0, x∈R
5x+5=5+2x+x²
x²-3x=0, x(x-3)=0, x=0 ∨ x=3
Otvet: 3
![\tt\displaystyle\sqrt{5x+5}=\sqrt{5+2x+x^{2}}\\\\5x + 5 = 5 + 2x + x^{2}\\\\x^{2}-3x=0\\\\x(x-3)=0\\\\x = 0~~~~~and~~~~~x - 3 = 0\\\\x = 0~~~~~and~~~~~x = 3](/tpl/images/0976/6086/548f9.png)
ответ3