Question 1192822
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{{{(sqrt(2x))/(sqrt(x)-sqrt(10))}}} is in the form {{{A/(B-C)}}}


{{{A = sqrt(2x)}}}


{{{B = sqrt(x)}}}


{{{C = sqrt(10)}}}


Multiply top and bottom by (B+C)
This way we have {{{(B-C)(B+C) = B^2 - C^2 = (sqrt(x))^2 - (sqrt(10))^2 = x - 10}}} in the denominator, thereby rationalizing it (i.e. removing the square root terms from the denominator). Refer to the difference of squares rule.


{{{A/(B-C)}}}


{{{(A(B+C))/((B-C)(B+C))}}}


{{{(A(B+C))/(B^2 - C^2)}}}


{{{(sqrt(2x)*(sqrt(x)+sqrt(10)))/(x-10)}}}


{{{(sqrt(2x)*sqrt(x)+sqrt(2x)*sqrt(10))/(x-10)}}}


{{{(sqrt(2x*x)+sqrt(2x*10))/(x-10)}}}


{{{(sqrt(2x^2)+sqrt(20x))/(x-10)}}}


{{{(sqrt(x^2*2)+sqrt(4*5x))/(x-10)}}}


{{{(sqrt(x^2)*sqrt(2)+sqrt(4)*sqrt(5x))/(x-10)}}}


{{{(x*sqrt(2)+2*sqrt(5x))/(x-10)}}}


Therefore,
{{{(sqrt(2x))/(sqrt(x)-sqrt(10)) = (x*sqrt(2)+2*sqrt(5x))/(x-10)}}}
such that {{{x >= 0}}} and {{{x <> 10}}}
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