Math, asked by nayesanbiram0, 1 year ago

Simplify the following into a fraction with rational denominator 1/(√6-√11+√5)

Answers

Answered by MVB
14
1/√5+√6-√11 = 1/(√a + √b + √c)

Therefore, 1/[(√a + √b) + √c]  = [(√a + √b) - √c] / { [(√a + √b) + √c].[(√a + √b) - √c] }

= [(√a + √b) - √c] / [(√a + √b)² - c]

= [(√a + √b) - √c] / (a + 2√ab + b - c)

= [(√a + √b) - √c] / [(a + b - c) + 2√ab]

= { [(√a + √b) - √c] * [(a + b - c) - 2√ab] } / { [(a + b - c) + 2√ab] * [(a + b - c) - 2√ab] }

= { [(√a + √b) - √c] * [(a + b - c) - 2√ab] } / [(a + b - c)² - 4ab]

→ in this case: a = 5; b = 6; c = 11

= { [(√5 + √6) - √11] * [(5 + 6 - 11) - 2√30] } / [(5 + 6 - 11)² - 120]

= { [(√5 + √6) - √11] * [- 2√30] } / [- 120]

= (√5 + √6 - √11) * 2√30 / 120

= (√5 + √6 - √11) * √30 / 60

= (√150 + √180 - √330) / 60 → Since: √150 = √(25 * 6) = 5√6

= (5√6 + √180 - √330) / 60 →
Since: √180 = √(36 * 5) = 6√5

= (5√6 + 6√5 - √330) / 60 is the solution.

Hope it helps!!!!

Answered by mindfulmaisel
12

Given:

\frac {1}{\sqrt {5} + \sqrt {6} - \sqrt {11}}

To find:

Simplify the fraction.

Solution:

\frac {1}{\sqrt {5} + \sqrt {6} - \sqrt {11}} = \frac {1}{\sqrt {a} + \sqrt {b} + \sqrt {c}}

Hence, \frac { 1 }{ { \sqrt { a } +\sqrt { b } }+\sqrt { c } } =\frac { 1 }{ { \sqrt { a } +\sqrt { b } }+\sqrt { c } } \times \frac { { \sqrt { a } +\sqrt { b } }-\sqrt { c } }{ { \sqrt { a } +\sqrt { b } }-\sqrt { c } }

= \frac { [ ( \sqrt { \mathrm { a } } + \sqrt { \mathrm { b } } ) - \sqrt { \mathrm { c } } ] } { \left[ ( \sqrt { \mathrm { a } } + \sqrt { \mathrm { b } } ) ^ { 2 } - \mathrm { c } \right] }

= \frac { [ ( \sqrt { a } + \sqrt { b } ) - \sqrt { c } ] } { ( a + 2 \sqrt { a b } + b - c ) }

= \frac { [ ( \sqrt { a } + \sqrt { b } ) - \sqrt { c } ] } { [ ( a + b - c ) + 2 \sqrt { a b } ] }

= \frac { \{ [ ( \sqrt { a } + \sqrt { b } ) - \sqrt { c } ] \times [ ( a + b - c ) - 2 \sqrt { a b } ] \} } { \{ [ ( a + b - c ) + 2 \sqrt { a b } ] \times [ ( a + b - c ) - 2 \sqrt { a b } ] \} }

= \frac { \{ [ ( \sqrt { a } + \sqrt { b } ) - \sqrt { c } ] \times [ ( a + b - c ) - 2 \sqrt { a b } ] \} } { \left[ ( a + b - c ) ^ { 2 } - 4 a b \right] }

Here a = 5; b = 6; c = 11

= \frac { \{ [ ( \sqrt { 5 } + \sqrt { 6 } ) - \sqrt { 11 } ] \times [ ( 5 + 6 - 11 ) - 2 \sqrt { 30 } ] \} } { \left[ ( 5 + 6 - 11 ) ^ { 2 } - 120 \right] }

= \frac { \{ [ ( \sqrt { 5 } + \sqrt { 6 } ) - \sqrt { 11 } ] \times [ - 2 \sqrt { 30 } ] \} } { [ - 120 ] }

= \frac { ( \sqrt { 5 } + \sqrt { 6 } - \sqrt { 11 } ) \times 2 \sqrt { 30 } } { 120 }

= \frac { ( \sqrt { 5 } + \sqrt { 6 } - \sqrt { 11 } ) \times \sqrt { 30 } } { 60 }

= \frac { ( \sqrt { 150 } + \sqrt { 180 } - \sqrt { 330 } ) } { 60 }

= \frac { ( 5 \sqrt { 6 } + \sqrt { 180 } - \sqrt { 330 } ) } { 60 }

As\quad [ \sqrt {180} = \sqrt {36 \times 5} = 6 \times \sqrt {5} ]

= \frac { ( 5 \sqrt { 6 } + 6 \sqrt { 5 } - \sqrt { 330 } ) } { 60 }

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