Precalculus math questions

1. Simplify the compound fractional expression.

(x/y)-(y-x) / (1/9x^2)-(1/9y^2)

2. Simplify the expression.

(1-x^2)^(1/2)+x^2(1-x^2)^(-1/2) / (1-x^2)

3. Find all real solutions of the equation.

(x+8)/(x-2) = (2)/(x+2) + (40)/(x^2-4)

4. Find all real solutions and check your answers.

(y + 3)^(2/3) − 2(y + 3)^(1/3) − 3 = 0

5. Find all real solutions and check your answers.

x^8 − 2x^4 − 35 = 0

1. (x^2 - y^2)/xy / ((9y^2 - 9x^2)/9x^2y^2))
= (x+y)(x-y)9x^2y^2 / (9(y-x)(y+x)xy)
= (x-y)xy / (y-x) = -xy

2. (1-x^2)^(1/2)+x^2(1-x^2)^(-1/2) / (1-x^2)
= (1-x^2)^(1/2)+x^2 / (1-x^2)^(1/2)(1-x^2)
= [(1-x^2)^(1/2)(1-x^2)^(1/2)(1-x^2) / (1-x^2)^(1/2)(1-x^2)] +x^2 / (1-x^2)^(1/2)(1-x^2)
= [(1-x^2)^2 / (1-x^2)^(1/2)(1-x^2)] +x^2 / (1-x^2)^(1/2)(1-x^2)
= [(1-x^2)^2 + x^2] / (1-x^2)^(1/2)(1-x^2)
= (x^4-x^2+1) / (1-x^2)^(3/2)

3. (x+8)(x+2)/(x-2)(x+2) = (2)(x-2)/(x+2)(x-2) + (40)/(x^2-4)
=> (x^2 + 10x +16) / (x^2-4) = (2x - 4)/(x^2-4) + 40/(x^2-4)
=> x^2 + 8x - 20 = 0
=> (x-2)(x+10) = 0
=> x = 2 or -10

However, if x=2 this will make denominator x^2 -4 be zero (0),
so only x = -10 is the answer.

4. let (y + 3)^(1/3) = Z, then you have
Z^2 - 2Z -3 = 0
(Z-3)(Z+1) = 0
Z = 3 or -1

(y + 3)^(1/3) = 3
=> y+3 = 27
=> y = 24

(y + 3)^(1/3) = -1
=> y+3 = -1
=> y = -4

5. x^8 − 2x^4 − 35
= (x^4-7)(x^4+5)
= (x^2- sqrt7)(x^2+sqrt7)(x^4+5)
= (x+ 7^(1/4))(x- 7^(1/4))(x^2+sqrt7)(x^4+5) = 0
so x=-7^(1/4) or x = 7^(1/4)
(x^2+sqrt7) and (x^4+5) can't be zero (0) since x^2 and x^4 are all positive real numbers.

Hope this helps.