Hi guys, I've been stuck on this question for days!
If x= √3/2
Find the value of the expression
1+x/1+√(1+x) + 1-x/1-√(1-x)
Thanks!
Maybe start by rationalising the denominator
$\displaystyle \displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}$
$\displaystyle \displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1-x}}{1-\sqrt{1-x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}$
What do you get?
following on from what u had....
(1+x)(1-√1-x)/)1+√1+x)(1-√1-x) + (1-x)(1+√1+x)/(1-√1-x)1+√1+x)
= (1+x)(1-√1-x) + (1-x)(1+√1+x)/1+√1+x)(1-√1-x)
=2-√1-x - x√1-x + √1+x -x1+x/(1+√1+x)(1-√1-x)
=2-√1-x(1+x) + √1+x(1-x)/(1+√1+x)(1-√1-x)
Thats as far as i could get :s
I don't think my suggestion will simplify it enough.
simplify ((1+(sqrt[3]/2) 1;/(1+sqrt[1+(sqrt[3&# 93;/2)]))+((1-(sqrt[3]/2))/(1-sqrt[1-(sqrt[3]/2)]
You'll get the most simplified form of it by using the hint pickslides has given you:
$\displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=\frac{(1+x)(1-\sqrt{1-x})+(1-x)(1+\sqrt{1+x})}{(1+\sqrt{1+x})(1-\sqrt{1-x})}$
$\displaystyle =\frac{1+x-\sqrt{1-x}-x\sqrt{1-x}+1-x+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}$
$\displaystyle =\frac{2-\sqrt{1-x}-x\sqrt{1-x}+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}$
Now you just have to substitute the given $\displaystyle x$ value in this expression which is not that easy.