rationalize 1 upon root 6 minus root 5 with a rational denominator
Answers
Step-by-step explanation:
Answer:
\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}
6
+
5
−
11
1
=
60
6
5
+5
6
+
330
Given: \frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}
6
+
5
−
11
1
We have to rationalize the denominator.
Consider,
\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}
(
6
+
5
)−
11
1
=\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}\times\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}=
(
6
+
5
)−
11
1
×
(
6
+
5
)−
11
(
6
+
5
)+
11
=\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5}-\sqrt{11})(\sqrt{6}+\sqrt{5}+\sqrt{11})}=
(
6
+
5
−
11
)(
6
+
5
+
11
)
(
6
+
5
)+
11
=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5})^2-(\sqrt{11})^2}=
(
6
+
5
)
2
−(
11
)
2
6
+
5
+
11
=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6})^2+(\sqrt{5})^2+2\sqrt{6\times5}-11}=
(
6
)
2
+(
5
)
2
+2
6×5
−11
6
+
5
+
11
=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}=
2
30
6
+
5
+
11
=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}\times\frac{\sqrt{30}}{\sqrt{30}}=
2
30
6
+
5
+
11
×
30
30
=\frac{(\sqrt{6}+\sqrt{5}+\sqrt{11})\times\sqrt{30}}{2(\sqrt{30})^2}=
2(
30
)
2
(
6
+
5
+
11
)×
30
=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{2\times30}=
2×30
6
5
+5
6
+
330
=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}=
60
6
5
+5
6
+
330
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