Question

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ANSWER BOTH QUESTION PLZ❤️❤️ ​

Answer 1

Answer:

1 ST ANSWER IS 1212

2 ND ANSWER IS 1341

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Step-by-step explanation:

Answer 2

$\bf{\underline{Questions}}$

1. Simplify $\sf{\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}} + \dfrac{2-\sqrt{3}}{2+\sqrt{3}}}$

2. Rationalise the denominator and evaluate $\sf{\dfrac{\sqrt{2}}{2+\sqrt{2}}}$ by taking $\sf{\sqrt{2}}$ = 1.414 upto three decimal places.

$\bf{\underline{Solutions}}$

1. $\sf{\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}} + \dfrac{2-\sqrt{3}}{2+\sqrt{3}}}$

= $\sf{\bigg(\dfrac{\sqrt{2} + \sqrt{3}}{2\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}\bigg) + \bigg(\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\bigg)}$

= $\sf{\bigg[\dfrac{(2+\sqrt{3})\times\sqrt{3}}{2\sqrt{3}\times\sqrt{3}}\bigg] + \bigg[\dfrac{(2-\sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}\bigg]}$

= $\sf{\dfrac{2\sqrt{3} + 3}{2\times3} + \bigg[\dfrac{(2)^2 - 2\times 2 \times\sqrt{3} + (\sqrt{3})^2}{4-3}\bigg]}$

= $\sf{\dfrac{2\sqrt{3} + 3}{6} + \bigg[\dfrac{4-4\sqrt{3} + 3}{1}\bigg]}$

= $\sf{\dfrac{2\sqrt{3} + 3}{6} + 7-4\sqrt{3}}$

= $\sf{\dfrac{2\sqrt{3} +3 + 42 - 24\sqrt{3}}{6}}$

= $\sf{\dfrac{-22\sqrt{3} + 45}{6}\:\:\boxed{\sf{Answer}}}$

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2. $\sf{\dfrac{\sqrt{2}}{2+\sqrt{2}}}$

= $\sf{\dfrac{\sqrt{2}}{2+\sqrt{2}}\times\dfrac{2-\sqrt{2}}{2-\sqrt{2}}}$

= $\sf{\dfrac{\sqrt{2}(2-\sqrt{2})}{(2)^2-(\sqrt{2})^2}}$

= $\sf{\dfrac{2\sqrt{2}-2}{4-2}}$

= $\sf{\dfrac{2\sqrt{2}-2}{2}}$

Taking 2 as common,

= $\sf{\dfrac{2(\sqrt{2}-1)}{2}}$

= $\sf{\sqrt{2} - 1}$

Taking $\sf{\sqrt{2} = 1.414}$

= $\sf{1.414-1}$

= $\sf{0.414\:\:\boxed{\sf{Answer}}}$

$\bf{\underline{Formulas\:used}}$

• $\sf{(a-b)^2 = a^2 - 2ab + b^2}$

• $\sf{a^2 - b^2 = (a+b)(a-b)}$