how( 1-underoot 2 / -1 = underoot 2 -1 )
$ how ( 2 underoot 5- 5 underoot 2 / -30 = 5 underoot 2 - 2 underoot 5 / 30 )

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Solution

$(a) . \frac{1 - \sqrt{2}}{- 1} Multiplying numerator and denominator by - 1, we get \frac{1 - \sqrt{2}}{- 1} = \frac{- 1 (1 - \sqrt{2})}{- 1 \times (- 1)} = \frac{- 1 + \sqrt{2}}{1} = \sqrt{2} - 1 (b) . \frac{2 \sqrt{5} - 5 \sqrt{2}}{- 30} = \frac{- 1 (2 \sqrt{5} - 5 \sqrt{2})}{- 1 \times (- 30)} = \frac{5 \sqrt{2} - 2 \sqrt{5}}{30}$

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Similar questions

The graph for the following system of equations :

$2 x + 3 y = 5$ and $6 x + 9 y = 40$ is shown

$S_{1} : \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

S2 : The two lines intersect each other.

How to rationalize -:

(a) 2−√3/2+√3
(b) 2√5+3√2/2√5−3√2

S_{1} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots \infty

S_{2} = \frac{1}{3} + \frac{1}{5^{2}} + \frac{1}{3^{3}} + \frac{1}{5^{4}} + \frac{1}{3^{5}} + \frac{1}{5^{6}} + \dots \infty If S_{1} - 2 S_{2} = \frac{a}{12}, then a =

N_{2} O_{5} = 2 {N O}_{2} + \frac{1}{2} O_{2}

30^{0}

N_{2} O_{5}

N_{2} O_{5}

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