Rationalising factor of $(\sqrt{x+y})$ is.

Hint:If the product of two surds is a rational number, then each surd is a rationalizing factor to others. Like if $\sqrt{2}$ is multiplied with $\sqrt{2}$, it will be $2$, which is rational number, so $\sqrt{2}$ is rationalizing factor of $\sqrt{2}$.

Complete step-by-step answer:
If a surd or surd with rational numbers present in the denominator of an equation, to simplify it or to omit the surds from the denominator, rationalization of surds is used. Surds are irrational numbers but if multiply a surd with a suitable factor, the result of multiplication will be a rational number.
In elementary algebra, root rationalisation is a process by which radicals in the denominator of an algebraic fraction are eliminated.
If the denominator is a monomial in some radical, say $a{{\sqrt{{{x}^{n}}}}^{k}}$ with k$<$n, rationalisation consists of multiplying the numerator and the denominator by \[{{\sqrt{x}}^{n-k}}\]and replacing ${{\sqrt{x}}^{n}}$ if $n$ is even or by $x$ if n is odd (if $k\ge n$, the same replacement allows us to reduce $k$ until it becomes lower than $n$.
This technique may be extended to any algebraic denominator, by multiplying the numerator and the denominator by all algebraic conjugates of the denominator, and expanding the new denominator into the norm of the old denominator. However, except in special cases, the resulting fractions may have huge numerators and denominators, and, therefore, the technique is generally used only in the above elementary cases.
This is the basic principle involved in rationalization of surds. The factor of multiplication by which rationalization is done, is called the rationalizing factor. If the product of two surds is a rational number, then each surd is a rationalizing factor to others. Like if $\sqrt{2}$ is multiplied with $\sqrt{2}$, it will be $2$, which is rational number, so $\sqrt{2}$ is rationalizing factor of $\sqrt{2}$.
Here we have to find the rationalizing factor of $(\sqrt{x+y})$.
So we get, the rationalizing factor of $(\sqrt{x+y})$is $(\sqrt{x+y})$.

Note:Read the question in a careful manner. You should know that the factor of multiplication by which rationalization is done, is called the rationalizing factor. Example we can say that $\sqrt{2}$ is a rationalizing factor of $\sqrt{2}$.The rationalising factor is not the one like conjugate which has a negative sign, it's the factor with the same sign.