finding x involving fractions and radicals

[gangnamstyle]

_____2_____ + _____1_____ __ _____1_____ =0
(x^2-36)^1/2 (x+6)^1/2 (x-6)^1/2

sorry i think i messed up showing the problem

2/sqrt(x^2-36) + 1/sqrt(x+6) - 1/sqrt(x-6) =0

ans:x=10


(x+7)^1/2 - (x+2)^1/2 = (x-1)^1/2 - (x-2)^1/2

ans:x=2


on the 2nd problem, im thinking of squaring both side of the equations to cancel out those sqrt but im not sure?

 

[pka]

_____2_____ + _____1_____ __ _____1_____ =0
(x^2-36)^1/2 (x+6)^1/2 (x-6)^1/2
2/sqrt(x^2-36) + 1/sqrt(x+6) - 1/sqrt(x-6) =0; ans=10

Note that is \(\displaystyle \displaystyle\frac{2+\sqrt{x-6}-\sqrt{x+6}}{\sqrt{x^2-36}}=0\)

 

[HallsofIvy]

It helps to know that \(\displaystyle (x- 6)(x+ 6)= x^2- 36\).

 

[lookagain]

_______2_______ . . . + . . . _______1_______ . . . __ . . . _______1_______ = 0
(x^2-36)^(1/2) . . . . . . . . . . (x+6)^(1/2) . . . . . . . . . . . . . (x-6)^(1/2)


sorry i think i messed up showing the problem

2/sqrt(x^2-36) + 1/sqrt(x+6) - 1/sqrt(x-6) = 0

ans:x=10


(x+7)^(1/2) - (x+2)^(1/2) = (x-1)^(1/2) - (x-2)^(1/2)

ans:x=2


on the 2nd problem, im thinking of squaring both side of the equations to cancel out those sqrt but im not sure?

gangnamstyle,

you have to use grouping symbols, such as parentheses, around your exponents
(and certain other expressions when they arise).