Question

$$\frac{1}{\sqrt{2}+\sqrt{5}+1}$$

Answer

$$(2*(-2*sqrt(2)+sqrt(5)-3+sqrt(10)))/(36-2^2*10)$$

Solution


Rationalize the denominator: \(\frac{1}{\sqrt{2}+\sqrt{5}+1} \cdot \frac{\sqrt{2}+\sqrt{5}-1}{\sqrt{2}+\sqrt{5}-1}=\frac{\sqrt{2}+\sqrt{5}-1}{2+\sqrt{10}-\sqrt{2}+\sqrt{10}+5-\sqrt{5}+\sqrt{2}+\sqrt{5}-1}\).
\[\frac{\sqrt{2}+\sqrt{5}-1}{2+\sqrt{10}-\sqrt{2}+\sqrt{10}+5-\sqrt{5}+\sqrt{2}+\sqrt{5}-1}\]
Collect like terms.
\[\frac{\sqrt{2}+\sqrt{5}-1}{(2+5-1)+(\sqrt{10}+\sqrt{10})+(-\sqrt{2}+\sqrt{2})+(-\sqrt{5}+\sqrt{5})}\]
Simplify  \((2+5-1)+(\sqrt{10}+\sqrt{10})+(-\sqrt{2}+\sqrt{2})+(-\sqrt{5}+\sqrt{5})\)  to  \(6+2\sqrt{10}\).
\[\frac{\sqrt{2}+\sqrt{5}-1}{6+2\sqrt{10}}\]
Rationalize the denominator: \(\frac{\sqrt{2}+\sqrt{5}-1}{6+2\sqrt{10}} \cdot \frac{6-2\sqrt{10}}{6-2\sqrt{10}}=\frac{6\sqrt{2}-4\sqrt{5}+6\sqrt{5}-10\sqrt{2}-6+2\sqrt{10}}{{6}^{2}-{(2\sqrt{10})}^{2}}\).
\[\frac{6\sqrt{2}-4\sqrt{5}+6\sqrt{5}-10\sqrt{2}-6+2\sqrt{10}}{{6}^{2}-{(2\sqrt{10})}^{2}}\]
Factor out the common term \(2\).
\[\frac{2(3\sqrt{2}-2\sqrt{5}+3\sqrt{5}-5\sqrt{2}-3+\sqrt{10})}{{6}^{2}-{(2\sqrt{10})}^{2}}\]
Collect like terms.
\[\frac{2((3\sqrt{2}-5\sqrt{2})+(-2\sqrt{5}+3\sqrt{5})-3+\sqrt{10})}{{6}^{2}-{(2\sqrt{10})}^{2}}\]
Simplify  \((3\sqrt{2}-5\sqrt{2})+(-2\sqrt{5}+3\sqrt{5})-3+\sqrt{10}\)  to  \(-2\sqrt{2}+\sqrt{5}-3+\sqrt{10}\).
\[\frac{2(-2\sqrt{2}+\sqrt{5}-3+\sqrt{10})}{{6}^{2}-{(2\sqrt{10})}^{2}}\]
Simplify  \({6}^{2}\)  to  \(36\).
\[\frac{2(-2\sqrt{2}+\sqrt{5}-3+\sqrt{10})}{36-{(2\sqrt{10})}^{2}}\]
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[\frac{2(-2\sqrt{2}+\sqrt{5}-3+\sqrt{10})}{36-{2}^{2}{\sqrt{10}}^{2}}\]
Use this rule: \({\sqrt{x}}^{2}=x\).
\[\frac{2(-2\sqrt{2}+\sqrt{5}-3+\sqrt{10})}{36-{2}^{2}\times 10}\]

Decimal Form: 0.215041