Question

(1) if root x-a/x-b + a/b = root x-b/x-a + b/a, b not equal to a, then value of x is

(2) if x = 2 root 24/root3+root2 then value of x+roo8/x-root8 + x+root12/x-root12

(3)if a = 2+root3/2-root3 & b = 2-root3/2+root3 then value of (a square + b square + ab) =

(4) if root 4x-9 + Root 4x+9 = 5+root7 then value of x is

(5) if x = 1/2+root3 , y = 1/2-root3 then the value of 1/x+1 + 1/y+1 = ?

please explain it brief

Answer 1

You must use brackets to correctly express your terms in the above.
Without them we can only guess what the correct order of operation is

e.g. in #4,
did you mean:
√(4x-9) + ...
or
√(4x) - 9 + ...
or, the way you typed it
(√4)(x) - 9 + ....

I will assume you meant:
√(4x-9) + √(4x+9) = 5 + √7
To solve this requires some rigorous algebra,
but by some lucky coincidence , we can match up the terms to get a solution.
let √(4x+9) = 5
4x+9 = 25
4x = 16
x = 4
and if we let √(4x-9) = √7
4x-9 = 7
4x=16 ----> x = 4
This "method" seldom works, I just happened to notice it.
if the right side had been .... = 6 + √7 it would not work .

#5 I will assume you meant
x = 1/(2+√3) and y = 1/(2-√3) instead of what you typed

rationalizing both I got
x = 2-√3 , and y = 2 + √3

then 1/(x+1) + 1/(y+1) ----> again I assumed that's what you meant to type
= 1/(2-√3 + 1) + 1/(2+√3 + 1)
= 1/(3-√3) + 1/(3+√3)
= [ (3+√3) + (3-√3) ] / ( (3-√3)(3+√3) )
= 6/(9-3)
= 6/6
= 1

Answer 2

thanks for explain it but still first 3 example pending please explain it i need it so badly

(1)√x-a/x-b + a/b = √x-b/x-a + b/a, b not equal to a, then value of x is

(2) if x = 2√24/√3+√2 then value of x+√8/x-√8 + x+√12/x-√12 .

(3) if a = 2+√3/2-√3 & b = 2-√3/2+√3 then value of (a square + b square + ab) =


Thanks

Answer 3

Solve kiya hua