Answer:
Step-by-step explanation:
[tex] \dfrac{7 + \sqrt{3} }{2 + \sqrt{3} } [/tex]
[tex] \\ [/tex]
rationalizing the fraction by (2-√3)
[tex] = \dfrac{7 + \sqrt{3} }{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } [/tex]
cross-multiplying
[tex] = \dfrac{(7 + \sqrt{3})(2 - \sqrt{3} ) }{(2 + \sqrt{3})(2 - \sqrt{3}) } [/tex]
opening the brackets and using identity
[tex] = \dfrac{14 - 7 \sqrt{3} + 2 \sqrt{3} - 3 }{ {(2)}^{2} - { (\sqrt{3} )}^{2} } [/tex]
[tex] = \dfrac{11 - 5 \sqrt{3} }{4 - 3} [/tex]
[tex] = \dfrac{11 - 5 \sqrt{3} }{1} [/tex]
[tex] = 11 - 5 \sqrt{3} [/tex]
[tex] \\ \\ [/tex]
IDENTITY USED:
[tex] \boxed{ (a-b)(a+b) = a^2 - b^2} [/tex]
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Verified answer
Answer:
(7+√3) / (2+√3) = (11-5√3)
Step-by-step explanation:
[tex] \dfrac{7 + \sqrt{3} }{2 + \sqrt{3} } [/tex]
[tex] \\ [/tex]
rationalizing the fraction by (2-√3)
[tex] = \dfrac{7 + \sqrt{3} }{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } [/tex]
[tex] \\ [/tex]
cross-multiplying
[tex] = \dfrac{(7 + \sqrt{3})(2 - \sqrt{3} ) }{(2 + \sqrt{3})(2 - \sqrt{3}) } [/tex]
[tex] \\ [/tex]
opening the brackets and using identity
[tex] = \dfrac{14 - 7 \sqrt{3} + 2 \sqrt{3} - 3 }{ {(2)}^{2} - { (\sqrt{3} )}^{2} } [/tex]
[tex] \\ [/tex]
[tex] = \dfrac{11 - 5 \sqrt{3} }{4 - 3} [/tex]
[tex] = \dfrac{11 - 5 \sqrt{3} }{1} [/tex]
[tex] = 11 - 5 \sqrt{3} [/tex]
[tex] \\ \\ [/tex]
IDENTITY USED:
[tex] \boxed{ (a-b)(a+b) = a^2 - b^2} [/tex]