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Answer:
[tex]\frac{5-\sqrt{2}}{27}[/tex]
Step-by-step explanation:
This kind of problem uses a sort of algebra trick.
To rationalize [tex]\frac{1}{5+\sqrt{2}}[/tex], multiply both numerator and denominator by what is called the conjugate radical -- the same expression but with opposite sign.
Note: this is really a clever name for 1, and multiplying by 1 does not change the value of an expression, it changes the way the expression looks.
Here goes:
[tex]\frac{1}{5+\sqrt{2}} \cdot \frac{5-\sqrt{2}}{5-\sqrt{2}}=\frac{5-\sqrt{2}}{25+5\sqrt{2}-5\sqrt{2}+2}\\=\frac{5-\sqrt{2}}{25+2}\\=\frac{5-\sqrt{2}}{27}[/tex]
See how the radical disappears from the denominator?
Step-by-step explanation:
Answer:
\frac{5-\sqrt{2}}{27}
27
5−
2
Step-by-step explanation:
This kind of problem uses a sort of algebra trick.
To rationalize \frac{1}{5+\sqrt{2}}
5+
2
1
, multiply both numerator and denominator by what is called the conjugate radical -- the same expression but with opposite sign.
Note: this is really a clever name for 1, and multiplying by 1 does not change the value of an expression, it changes the way the expression looks.
Here goes:
\begin{gathered}\frac{1}{5+\sqrt{2}} \cdot \frac{5-\sqrt{2}}{5-\sqrt{2}}=\frac{5-\sqrt{2}}{25+5\sqrt{2}-5\sqrt{2}+2}\\=\frac{5-\sqrt{2}}{25+2}\\=\frac{5-\sqrt{2}}{27}\end{gathered}
5+
2
1
⋅
5−
2
5−
2
=
25+5
2
−5
2
+2
5−
2
=
25+2
5−
2
=
27
5−
2
See how the radical disappears from the denominator?