Method isn't full proof. If x=4 y=5 then you'd mistakenly get answer A?
rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)
Source: Total GMAT Math - Jeff Sackmann
I. Algebraic
Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)
\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)
II. Substitute some value for x and y.
Take x as 9 and y as 16..
\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)
(A)\(\frac{1}{y}=\frac{1}{16}\)....No
(B)\(\sqrt{2x + y}\).....Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)...No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\).....Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)...Yes
E