Question

1 divied by (√2+√3)-√4.​

Answer 1

1 divided by (√2+√3) - √4

1 ÷ (√5)-√4

1 ÷ √5-√4

1 ÷ √1

1 ÷ 1

1

So final answer is 1.

Answer 2

To find:

$\tt\longrightarrow\dfrac1{(\sqrt2+\sqrt3)-\sqrt4}$

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Solution:

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>> We will solve this problem by the method of rationalisation, where we multiply both numerator and denominator by the conjugate of the denominator.

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[Conjugate basically means the changing of sign between two terms.]

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:: Conjugate of (√2+√3)-√4 will be (√2+√3)+√4.

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$\tt\longrightarrow\dfrac{1}{(\sqrt2+\sqrt3)-\sqrt4}\times\Bigg[\dfrac{(\sqrt 2+\sqrt3)+\sqrt4}{(\sqrt2+\sqrt3)+\sqrt4}\Bigg]$

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big[\Big(\sqrt2+\sqrt3\Big)-\sqrt4\Big]\Big[\Big(\sqrt2+\sqrt3\Big)+\sqrt4\Big]}}$

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:: Now applying identity (A+B)(A-B)=A²-B² in denominator where A=(√2+√3) and B=√4.

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big[\Big(\sqrt2+\sqrt3\Big)^2-\Big(\sqrt4\Big)^2\Big]}}$

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big(\sqrt2+\sqrt3\Big)^2-\Big(4\Big)}}$

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:: Now applying identity (A+B)²=A²+B²+2AB in denominator where A=√2 and B=√3.

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big(\sqrt2\Big)^2+\Big(\sqrt3\Big)^2+2\Big(\sqrt 2\Big)\Big(\sqrt3\Big)-\Big(4\Big)}}$

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:: Now solving it

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big(2\Big)+\Big(3\Big)+2\Big(\sqrt 6\Big)-\Big(4\Big)}}$

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big(2\sqrt 6\Big)+\Big(1\Big)}}$

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{2\sqrt 6+1}}$

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>> Now again rationalising the denominator by multiplying both numerator and denominator with the conjugate of its denominator.

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:: Now conjugate for (2√6+1) will be (2√6-1), multiply it with both numerator and denominator.

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$\tt\longrightarrow{\dfrac{\Big(\sqrt 2+\sqrt3\Big)+\sqrt4}{\Big ( 2\sqrt 6+1\Big)}\times \dfrac{\Big(2\sqrt6-1\Big)}{\Big(2\sqrt6-1\Big)}}$

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$\tt\longrightarrow\dfrac{\Big(\sqrt 2+\sqrt3+\sqrt4\Big)\Big(2\sqrt6-1\Big)}{\Big(2\sqrt6-1\Big)\Big ( 2\sqrt 6+1\Big)}$

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:: Applying identity in denominator (A+B)(A-B)=A²-B², where A=2√6 and B=1

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$\tt\longrightarrow\dfrac{\Big(\sqrt 2+\sqrt3+\sqrt4\Big)\Big(2\sqrt6-1\Big)}{\Big(2\sqrt6\Big)^2-\Big(1\Big)^2}$

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$\tt\longrightarrow\dfrac{2\sqrt6\Big(\sqrt 2+\sqrt3+\sqrt4\Big)-1\Big(\sqrt 2+\sqrt3+\sqrt4\Big)}{\Big(2\sqrt6\Big)^2-\Big(1\Big)^2}$

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$\tt\longrightarrow\dfrac{\Big(2\sqrt {12}+2\sqrt{18}+4\sqrt6\Big)+\Big(-\sqrt 2-\sqrt3-\sqrt4\Big)}{24-1}$

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$\tt\longrightarrow\dfrac{4\sqrt {3}+6\sqrt{2}+4\sqrt6-\sqrt 2-\sqrt3-2}{23}$

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$\tt\longrightarrow\dfrac{3\sqrt {3}+5\sqrt{2}+4\sqrt6-2}{23}$

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Since it can't be simplified any more, this is your answer.