Question

1/4+√2+√5, rationalise the denominator *

Answer 1

Answer:

this is the solution of the Question.

Answer 2

$5\sqrt{2} +6-\sqrt{10}-\sqrt{5}$

Step-by-step explanation:

• Irrational numbers are those numbers which cannot be expressed as the fraction in the form of p/q, where q is a non-zero number.
• Here, according to the given information, the expression is,

    1/4+√2+√5

Now,

$\frac{1}{4+\sqrt{2}+\sqrt{5} }$ to be free from irrational denominators, we need to use the help of  the well - known identity that is (a+b)(a-b) = a²-b².

Then, we have,

$\frac{1}{4+\sqrt{2}+\sqrt{5} }\\=\frac{(4+\sqrt{2}-\sqrt{5} )}{(4+\sqrt{2}+\sqrt{5})(4+\sqrt{2}-\sqrt{5} )}\\=\frac{(4+\sqrt{2}-\sqrt{5} )}{\sqrt{2}-1 }$

Again, we have an irrational denominator. To resolve this as well, we need to use the same identity that is (a+b)(a-b) = a²-b², and get,

$\frac{(4+\sqrt{2}-\sqrt{5} )(\sqrt{2}+1) }{(\sqrt{2}+1)\sqrt{2}-1)} \\=\frac{(4+\sqrt{2}-\sqrt{5} )(\sqrt{2}+1) }{2-1} \\\\=(4+\sqrt{2}-\sqrt{5})(\sqrt{2}+1)\\=5\sqrt{2} +6-\sqrt{10}-\sqrt{5}$

Thus, the answer is $5\sqrt{2} +6-\sqrt{10}-\sqrt{5}$.

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