Simplify 4 divide by
2+√3+√7
by rationalizing the denominator.
Answers
Step-by-step explanation:
Answer:
\frac{4}{2+\sqrt{3}+\sqrt{7}}
2+
3
+
7
4
= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}
3
(2
3
+3−
21
)
Explanation:
Given \frac{4}{2+\sqrt{3}+\sqrt{7}}
2+
3
+
7
4
multiply numerator and denominator by (2+√3-√7), we get
=\frac{4(2+\sqrt{3}-\sqrt7)}{[(2+\sqrt{3})+\sqrt{7}][(2+\sqrt{3})-\sqrt{3}]}
[(2+
3
)+
7
][(2+
3
)−
3
]
4(2+
3
−
7
)
= \frac{4(2+\sqrt{3}-\sqrt{7})}{(2+\sqrt{3})^{2}-(\sqrt{7})^{2}}
(2+
3
)
2
−(
7
)
2
4(2+
3
−
7
)
= \frac{4(2+\sqrt{3}-\sqrt{7})}{2^{2}+2\times2\times\sqrt{3}+(\sqrt{3})^{2}-7}
2
2
+2×2×
3
+(
3
)
2
−7
4(2+
3
−
7
)
=\frac{4(2+\sqrt{3}-\sqrt{7})}{4+4\sqrt{3}+3-7}
4+4
3
+3−7
4(2+
3
−
7
)
=\frac{4(2+\sqrt{3}-\sqrt{7})}{4\sqrt{3}}
4
3
4(2+
3
−
7
)
=\frac{(2+\sqrt{3}-\sqrt{7})}{\sqrt{3}}
3
(2+
3
−
7
)
Rationalising the denominator, we get
= \frac{(2+\sqrt{3}-\sqrt{7})\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}
3
×
3
(2+
3
−
7
)×
3
= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}
3
(2
3
+3−
21
)
Therefore,
\frac{4}{2+\sqrt{3}+\sqrt{7}}
2+
3
+
7
4
= \frac{(2\sqrt{3}+3-\sqrt{21})}{3}
3
(2
3
+3−
21
)