Question

Simplify..... And I will mark as a brainlist ​

Answer 1

Solution

1) $\sf 4\sqrt{75}+\sqrt{27}-\sqrt{3}$

Write this way also

√75 = 4√3 × 5 × 5 = 20√3

√27 = √3 × 3 × 3 = 3√3

20√3 + 3√3 - √3

Take √3 as a common

= √3(20 + 3 -1)

= 22√3

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2) $\sf \sqrt{50}-\sqrt{72}-\sqrt{18}$

Write in this way also

√50 = √5 × 5 × 2 = 5√2

√72 = √2 × 2 × 2 × 3 × 3 = 2×3√2 = 6√2

√18 = √2 × 3 × 3 = 3√2

5√2 - 6√2 - 3√2

Take √2 as a common

= √2 (5 - 6 - 3)

= -4√2

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$\sf 3)\;\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{3}-\sqrt{2}} \\ \\ \sf =\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} \\ \\ \sf =\dfrac{\sqrt{3}+\sqrt{3}}{3-2} \\ \\ \sf =\sqrt{3}(1+1) = 2\sqrt{3}$

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4) $\sf If\:x=2+\sqrt{3}\:then\:find\:x+\dfrac{1}{x}$

$\implies\sf x =2+\sqrt{3}$

$\sf \dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}} \\ \\ \sf \dfrac{1}{x}=\dfrac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2} \\ \\ \sf \dfrac{1}{x}=\dfrac{2-\sqrt{3}}{2-3}=2-\sqrt{3}$

$\sf So,\;x+\dfrac{1}{x}$

putting the value of x and 1/x

= 2+√3 + 2-√3 = 4

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5) $\sf If\:x=3+2\sqrt{2}\:then\:find\:x-\frac{1}{x}$

$\implies\sf x =3+2\sqrt{2}$

$\sf \dfrac{1}{x}=\dfrac{1}{3+3\sqrt{3}}\times\dfrac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ \\ \sf \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{(3)^2-(2\sqrt{3})^2} \\ \\ \sf \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}$

$\sf So,\;x-\dfrac{1}{x}$

Putting the value of x and 1/x

= 3 + 2√2 - (3 - 2√2)

= 3 + 2√2 - 3 + 2√2

= 2√2 + 2√2

take √2 as a common

= √2(2 + 2)

= 4√2