√х+2=2 +√х-6 √d+20=3√d-1

Question

√х+2=2 +√х-6
√d+20=3√d-1

Answer 1

1)ОДЗ x+2≥0 U x-6≥0⇒x≥-2 U x≥6⇒x∈[6;∞)
x+2=4+4√(x-6)+x-6
4√(x-6)=4
√(x-6)=1
x-6=1
x=6+1=7
2)ОДЗ  d+20≥0 U d-1≥0⇒d≥-20 U d≥1⇒d∈[1;∞)
d+20=9(d-1)
d+20=9d-9
9d-d=20+9
8d=29
d=29/8= 3 5/8