ОДЗ
x+4≥0⇒x≥-4
x-1≥0⇒x≥1
x-4≥0⇒x≥4
x∈[4;∞)
x+4-2√(x²+3x-4)+x-1=x-4
2√(x²+3x-4)=x+7
4x²+12x-16-x²-14x-49=0
3x²-2x-65=0
D=4+780=784
x1=(2-28)/6=-13/3∉ОДЗ
х2=(2+28)/6=5
ОДЗ
x+4≥0⇒x≥-4
x-1≥0⇒x≥1
x-4≥0⇒x≥4
x∈[4;∞)
x+4-2√(x²+3x-4)+x-1=x-4
2√(x²+3x-4)=x+7
4x²+12x-16-x²-14x-49=0
3x²-2x-65=0
D=4+780=784
x1=(2-28)/6=-13/3∉ОДЗ
х2=(2+28)/6=5