Question

Помогите! Алгебра 10 класс!!! Корни!!! Сколько сможете!

Answer 1

58) (x-y)/(√x+√y)-(√y-y)/√y=(√x+√y)(√x-√y)/(√x+√y)-√y(1-√y)/√y=(√x-√y)-1+√y=
=√x-1

60)
(√x-√y)/(⁴√x-⁴√y)-(⁴√(xy)+√y)/(⁴√x+⁴√y)=
=(⁴√x-⁴√y)(⁴√x+⁴√y)/(⁴√x-⁴√y)-⁴√y(⁴√x+⁴√y)/(⁴√x+⁴√y)=(⁴√x+⁴√y)-⁴√y=⁴√x

62) [√a/(b+√(ab))-√a/(b-√(ab))]·(b²-ab)/(a√b)=

= [(√a/√b)(1/(√b+√a)-1/(√b-√a)]·(b(b-a))/(a√b)=

=[(√a/√b)/(a√b)]·{[(√b-√a)-(√b+√a)]/(b-a)}·(b(b-a))=
=[(1)/(b√a)]·{[(-2√a)]}·b=-2

64)  (a-b)/(a^(1/2)-b^(1/2))- { (a^(3/2)-b^(3/2))/(a+a^(1/2)b^(1/2)+b)}=

(a^(1/2)+b^(1/2))-
-{ [(a^(1/2)-b^(1/2))·(a+a^(1/2)b^(1/2)+b)}/(a+a^(1/2)b^(1/2)+b)}=
=(a^(1/2)+b^(1/2))-(a^(1/2)-b^(1/2))=2b^(1/2)=2√b

66)  [(x-1)/(x+√x+1) : [(√x+1)/(x^(1,5)-1)] +2/[(√x)^(-1]=
((√x-1)(√x+1)/(x+√x+1) ·(x^(1/2)-1)·(x+√x+1) /(√x+1)+2(√x)=(√x-1)²+2√x=
=x-2√x+1+2√x=x+1

Answer 2

Огромное спасибо!!!!)

Answer 3

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