Question

$$\frac{ 1 }{ 3- \sqrt{ 8 } } -2 \sqrt{ 2 } +6$$

Answer

9

Solution


Simplify  \(\sqrt{8}\)  to  \(2\sqrt{2}\).
\[\frac{1}{3-2\sqrt{2}}-2\sqrt{2}+6\]
Rationalize the denominator: \(\frac{1}{3-2\sqrt{2}} \cdot \frac{3+2\sqrt{2}}{3+2\sqrt{2}}=\frac{3+2\sqrt{2}}{{3}^{2}-{(2\sqrt{2})}^{2}}\).
\[\frac{3+2\sqrt{2}}{{3}^{2}-{(2\sqrt{2})}^{2}}-2\sqrt{2}+6\]
Simplify  \({3}^{2}\)  to  \(9\).
\[\frac{3+2\sqrt{2}}{9-{(2\sqrt{2})}^{2}}-2\sqrt{2}+6\]
Rationalize the denominator: \(\frac{3+2\sqrt{2}}{9-{(2\sqrt{2})}^{2}} \cdot \frac{9+{(2\sqrt{2})}^{2}}{9+{(2\sqrt{2})}^{2}}=\frac{27+24+18\sqrt{2}+16\sqrt{2}}{{9}^{2}-{({(2\sqrt{2})}^{2})}^{2}}\).
\[\frac{27+24+18\sqrt{2}+16\sqrt{2}}{{9}^{2}-{({(2\sqrt{2})}^{2})}^{2}}-2\sqrt{2}+6\]
Collect like terms.
\[\frac{(27+24)+(18\sqrt{2}+16\sqrt{2})}{{9}^{2}-{({(2\sqrt{2})}^{2})}^{2}}-2\sqrt{2}+6\]
Simplify  \((27+24)+(18\sqrt{2}+16\sqrt{2})\)  to  \(51+34\sqrt{2}\).
\[\frac{51+34\sqrt{2}}{{9}^{2}-{({(2\sqrt{2})}^{2})}^{2}}-2\sqrt{2}+6\]
Factor out the common term \(17\).
\[\frac{17(3+2\sqrt{2})}{{9}^{2}-{({(2\sqrt{2})}^{2})}^{2}}-2\sqrt{2}+6\]
Simplify  \({9}^{2}\)  to  \(81\).
\[\frac{17(3+2\sqrt{2})}{81-{({(2\sqrt{2})}^{2})}^{2}}-2\sqrt{2}+6\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[\frac{17(3+2\sqrt{2})}{81-{(2\sqrt{2})}^{4}}-2\sqrt{2}+6\]
Rewrite \(81-{(2\sqrt{2})}^{4}\) in the form \({a}^{2}-{b}^{2}\), where \(a=9\) and \(b=8\).
\[\frac{17(3+2\sqrt{2})}{{9}^{2}-{8}^{2}}-2\sqrt{2}+6\]
Simplify  \({9}^{2}\)  to  \(81\).
\[\frac{17(3+2\sqrt{2})}{81-{8}^{2}}-2\sqrt{2}+6\]
Simplify  \({8}^{2}\)  to  \(64\).
\[\frac{17(3+2\sqrt{2})}{81-64}-2\sqrt{2}+6\]
Simplify  \(81-64\)  to  \(17\).
\[\frac{17(3+2\sqrt{2})}{17}-2\sqrt{2}+6\]
Cancel \(17\).
\[3+2\sqrt{2}-2\sqrt{2}+6\]
Collect like terms.
\[(3+6)+(2\sqrt{2}-2\sqrt{2})\]
Simplify.
\[9\]