Question

1) Simplify :-
4√3/2 - √2 - 30/4√2 - √18 - √18/3 - √12
2) If (1/x - 1/y) ∝ 1/x - y, then show that (x² + y²) ∝ xy​

Answer 1

Q1}

This problem is rather simpler:) You may calculate term wise by rationalising the denominator & then combine them. That'll be systematic….

[1] = (4√3) / (2-√2) = (4√3)(2+√2) / (2-√2)(2+√2)

= (8√3+4√6) /2

= 4√3+2√6 …………………(1)

[2]= 30 / (4√3-√18)

= 30*(4√3+√18) /(4√3-√18)(4√3+√18)

=( 120√3 +30√18) /(48–18)

= 30(4√3+√18) /30

= 4√3 + 3√2 ………………(2)

[3]= √18 / (3–2√3)

= √18(3+2√3) / (3–2√3) (3+2√3)

= (3√18+2√54) /9–12

= (9√2 + 6√6) /-3

= -3√2 -2√6 ………………(3)

Now by combining all 3 terms:

(4√3+2√6) — (4√3+3√2) — ( -3√2 -2√6)

= 4√3+2√6–4√3–3√2+3√2+2√6

= 2√6 + 2√6

= 4√6 ………….ANS

See the attachment for question 1)

Q2}

1/y-1/x∝1/x-y

Or (x-y)/xy=k×1/(x-y)(where k≠0)

Or (x-y)²=kxy

Or x²-2xy+y²=kxy

Or x²+y²=(k+2)xy

Or (x²+y²)/xy=k+2

Or x/y+y/x=k+2

Or (x/y)²+1=c(x/y) (where c=k+2 =constant)

Or (x/y)²-2(x/y)(c/2)+c²/4+1-c/4=0

Or (x/y-c/2)²=(c²/4)–1

Or x/y-c/2=±√(c²-4)/2

Or x/y=(c/2)±√(c²-4)/2

Or x/y=constant

$hope \: you \: like \: it.$

Answer 2

Answer:

2-√2)(2+√2)

= (8√3+4√6) /2

= 4√3+2√6 …………………(1)

[2]= 30 / (4√3-√18)

= 30*(4√3+√18) /(4√3-√18)(4√3+√18)

=( 120√3 +30√18) /(48–18)

= 30(4√3+√18) /30

= 4√3 + 3√2 ………………(2)

[3]= √18 / (3–2√3)

= √18(3+2√3) / (3–2√3) (3+2√3)

= (3√18+2√54) /9–12

= (9√2 + 6√6) /-4

=-3√2 -2√6 ………………(3)

Now by combining all 3 terms:

(4√3+2√6) — (4√3+3√2) — ( -3√2 -2√6)

= 4√3+2√6–4√3–3√2+3√2+2√6

= 2√6 + 2√6

= 4√6 ………….ANS

}

1/y-1/x∝1/x-y

Or (x-y)/xy=k×1/(x-y)(where k≠0)

Or (x-y)²=kxy

Or x²-2xy+y²=kxy

Or x²+y²=(k+2)xy

Or (x²+y²)/xy=k+2

Or x/y+y/x=k+2

Or (x/y)²+1=c(x/y) (where c=k+2 =constant)

Or (x/y)²-2(x/y)(c/2)+c²/4+1-c/4=0

Or (x/y-c/2)²=(c²/4)–1

Or x/y-c/2=±√(c²-4)/2

Or x/y=(c/2)±√(c²-4)/2

Or x/y=constant