Question

Simplify by rationalising the denomination of 7√3-5√2/√48+√18​

Answer 1

Step-by-step explanation:

Given \: \frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{48}+\sqrt{18})}Given(48+18)(73−52)

=\frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{4\times 4\times 3}+\sqrt{3\times 3\times 2})}(4×4×3+3×3×2)(73−52)

=\frac{(7\sqrt{3}-5\sqrt{2})}{(4\sqrt{3}+3\sqrt{2})}(43+32)(73−52)

/* Multiply numerator and denominator by (4\sqrt{3}-3\sqrt{2})(43−32) , we get

=\frac{(7\sqrt{3}-5\sqrt{2})(4\sqrt{3}-3\sqrt{2})}{(4\sqrt{3}+3\sqrt{2})(4\sqrt{3}-3\sqrt{2})}(43+32)(43−32)(73−52)(43−32)

=\frac{28\times 3-21\sqrt{6}-20\sqrt{6}+15\times 2}{(4\sqrt{3})^{2}-(3\sqrt{2})^{2}}(43)2−(32)228×3−216−206+15×2

=\frac{84-41\sqrt{6}+30}{48-18}48−1884−416+30

=\frac{114-41\sqrt{6}}{30}30114−416

/* Denominator rationalised .

Therefore,

\begin{gathered} \frac{(7\sqrt{3}-5\sqrt{2})}{(\sqrt{48}+\sqrt{18})}\\=\frac{114-41\sqrt{6}}{30}\end{gathered}(48+18)(73−52)=30114−416

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Answer 2

Answer:

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