Question

Hello‼️

Solve it by rationalization..❗

Answer:= (-1.466)

So solve ...✅

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Answer 1

${\boxed { \tt solution \ratio - }}$

Hlo Neha♡

here is ur answer : )

$\tt \frac{1 }{ \sqrt{3} - \sqrt{2} - 1 } = \frac{1}{ \sqrt{3} ( \sqrt{2} + 1) } \times \frac{ \sqrt{3} + (\sqrt{2} + 1) }{ \sqrt{3} + \sqrt{(2} + 1)} \: \: [rationalise ]$

$\tt = \frac{ \sqrt{3} + \sqrt{2} + 1}{ (\sqrt{3 {)}^{2} } - (\sqrt{2} + 1 {)}^{2} } \: \: [ \therefore (a - b)(a + b) = {a}^{2} - {b}^{2} ]$

$\tt = \frac{ \sqrt{3} + \sqrt{2} + 1}{3 -(2 + 1 + 2 \sqrt{2} ) }$

$\tt = \frac{ \sqrt{3} + \sqrt{2} + 1}{ - 2 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

$\tt = \frac{ \sqrt{6} + 2 + \sqrt{2} }{ -4 }$

$\tt = \frac{ - (2.499 + 2 + 1.414) }{ 4 }$

$\tt = \frac{ - 5.863 }{ 4 }$

$\boxed{ \tt = - 1.466}$

Answer 2

$\sf{\displaystyle\frac{1}{\sqrt{3}-\sqrt{2}-1}}$

We need to rationalize the denominator by removing radicals from it. In our example, the denominator contains 2 radical terms.

It needs to be multiplied by an expression that completes the product on the left side of the difference of two squares rule : $\sf{A^{2}-B^{2}=(A-B)(A+B)}$

$\to\sf{\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}-1} \cdot \frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3}-\sqrt{2}+1}}$

$\to\sf{\displaystyle\frac{(\sqrt{3}-\sqrt{2})+1}{(\sqrt{3}-\sqrt{2}-1)((\sqrt{3}-\sqrt{2})+1)}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{(\sqrt{3}-\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)}}$

We need to expand this term by multiplying two expressions. The following product distributive property will be used : $\sf{(A+B)(C+D)=A C+A D+B C+B D}$ .

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3}\cdot \sqrt{3}-\sqrt{3}\:\cdot \:\sqrt{2}+\sqrt{3}-\sqrt{2}\:\cdot \:\sqrt{3}+\sqrt{2}\:\cdot \:\sqrt{2}-\sqrt{2}}}$

We need to combine all radicands in this term, under one radical sign. This is possible, because all the indexes are equal .

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3\:\cdot \:3}-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{2\:\cdot \:3}+\sqrt{2\:\cdot \:2}-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

We need to combine like factors in this term by adding up all the exponents and copying the base. No exponent implies the value of 1.

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3^{1+1}}-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{2\:\cdot \:3}+\sqrt{2^{1+1}}-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3^2}-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{2\:\cdot \:3}+\sqrt{2^2}-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

We can simplify a radical by removing the GCF between the exponent in the radicand and the radical index. In our example, the exponent in the radicand is equal to 2 , the radical index is equal to 2 and the greatest common factor is equal to 2. Since the GCF is equal to the radical index, we can completely remove the radical sign.

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{3^{\frac{2}{2}}-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{2\:\cdot \:3}+2^{\frac{2}{2}}-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

Reduce this fraction to the lowest terms.  This can be done by dividing out those factors that appear both in the numerator and in the denominator.

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{3-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{2\:\cdot \:3}+2-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{3-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{3\cdot 2}+2-\sqrt{2}-\sqrt{3}+\sqrt{2}-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-\sqrt{3\:\cdot \:2}-\sqrt{3\:\cdot \:2}+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}+3+2-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-2\sqrt{3\cdot \:2}+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}+3+2-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-2\sqrt{3\cdot \:2}+\sqrt{3}-\sqrt{3}+3+2-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-2\sqrt{3\cdot \:2}+3+2-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-2\sqrt{6}+3+2-1}}$

$\to\sf{\displaystyle\frac{\sqrt{3}-\sqrt{2}+1}{-2\sqrt{6}+4}}$

Repeat This Step : We need to rationalize the denominator by removing radicals from it .........................

$\to\sf{\displaystyle\frac{\left(\sqrt{3}-\sqrt{2}+1\right)\left(4+2\sqrt{6}\right)}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{\sqrt{3}\cdot \:4+\sqrt{3}\cdot \:2\sqrt{6}+\left(-\sqrt{2}\right)\cdot \:4+\left(-\sqrt{2}\right)\cdot \:2\sqrt{6}+1\cdot \:4+1\cdot \:2\sqrt{6}}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{4\sqrt{3}+2\sqrt{3}\sqrt{6}-4\sqrt{2}-2\sqrt{2}\sqrt{6}+1\cdot \:4+1\cdot \:2\sqrt{6}}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{4\sqrt{3}+6\sqrt{2}-4\sqrt{2}-4\sqrt{3}+4+2\sqrt{6}}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{4\sqrt{3}+2\sqrt{2}-4\sqrt{3}+4+2\sqrt{6}}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{2\sqrt{2}+4+2\sqrt{6}}{\left(4-2\sqrt{6}\right)\left(4+2\sqrt{6}\right)}}$

$\to\sf{\displaystyle\frac{2\sqrt{2}+4+2\sqrt{6}}{4^2-\left(2\sqrt{6}\right)^2}}$

$\to\sf{\displaystyle\frac{2\sqrt{2}+4+2\sqrt{6}}{16-24}}$

$\to\sf{\displaystyle\frac{2\sqrt{2}+4+2\sqrt{6}}{-8}}$

$\to\sf{-\displaystyle\frac{2\sqrt{2}+4+2\sqrt{6}}{8}}$

$\to\sf{\displaystyle\frac{2\left(\sqrt{2}+2+\sqrt{6}\right)}{8}}$

$\to\sf{\displaystyle\frac{\sqrt{2}+2+\sqrt{6}}{4}}$

Given : $\sf{\sqrt{2}}$ = 1.414 and $\sf{\sqrt{6}}$ = 2.449

$\to\sf{\displaystyle\frac{1.414+2+2.449}{4}}$

$\to\sf{\displaystyle\frac{5.863}{4}}$

$\large\boxed{\boxed{\to\sf{-1.466}}}$