If x > 0, then 1/[√(2x)+√x] =
A. 1/√(3x)
B. 1/[2√(2x)]
C. 1/(x√2)
D. (√2-1)/√x
E. (1+√2)/√x
A. 1/√(3x)
B. 1/[2√(2x)]
C. 1/(x√2)
D. (√2-1)/√x
E. (1+√2)/√x
It is a problem of rationalization of denominator of a irrational denominator. Which can be done by using the formula (a + b)(a - b) = (a^2 - b^2)nickhar130 wrote:If x > 0, then 1/[√(2x)+√x] =
A. 1/√(3x)
B. 1/[2√(2x)]
C. 1/(x√2)
D. (√2-1)/√x
E. (1+√2)/√x
... 1/[√(2x) + √x]
= [√(2x) - √x]/[(√(2x) + √x)*(√(2x) - √x)] ....... Multiply with [√(2x) - √x]
= [√(2x) - √x]/[2x - x]
= [√(2x) - √x]/x
= [√2 - 1]/√x ...................................................... Divide with √x
The correct answer is D.
