Rationalize the denominator a^2÷square root of a^2+b^2 +b

1.	Rationalize the denominator a^2÷square root of a^2+b^2 +b
Answer» SOLUTION TO DETERMINE Rationalize the denominator $\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} }$ EVALUATION $\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} }$ $= \displaystyle \sf{ \frac{ {a}^{2} { \sqrt{ ({a}^{2} + {b}^{2} } - b)}}{ (\sqrt{ {a}^{2} + {b}^{2} } + b){ ( \sqrt{ {a}^{2} + {b}^{2} } - b)}} }$ $= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {( \sqrt{ {a}^{2} + {b}^{2} } )}^{2} - {b}^{2} } }$ $= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {a}^{2} + {b}^{2} - {b}^{2} } }$ $= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {a}^{2} } }$ $= \displaystyle \sf{ \sqrt{ {a}^{2} + {b}^{2} } - b }$ FINAL ANSWER $\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} = \sqrt{ {a}^{2} + {b}^{2} } - b}$ ━━━━━━━━━━━━━━━━ Learn more from Brainly :- 1. If x=√3a+2b + √3a-2b / √3a+2b - √3a-2b prove that bx²-3ax+b=0 brainly.in/question/19664646 2. if the real POSITIVE FOURTH root of(28+16√3) is divided by (2+√3) and the result is EXPRESSED as (a+b√3) with a&b both brainly.in/question/26773744

Answer»

TO DETERMINE

Rationalize the denominator

$\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} }$

EVALUATION

$\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} }$

$= \displaystyle \sf{ \frac{ {a}^{2} { \sqrt{ ({a}^{2} + {b}^{2} } - b)}}{ (\sqrt{ {a}^{2} + {b}^{2} } + b){ ( \sqrt{ {a}^{2} + {b}^{2} } - b)}} }$

$= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {( \sqrt{ {a}^{2} + {b}^{2} } )}^{2} - {b}^{2} } }$

$= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {a}^{2} + {b}^{2} - {b}^{2} } }$

$= \displaystyle \sf{ \frac{{a}^{2} ({ \sqrt{ {a}^{2} + {b}^{2} } - b)}}{ {a}^{2} } }$

$= \displaystyle \sf{ \sqrt{ {a}^{2} + {b}^{2} } - b }$

FINAL ANSWER

$\displaystyle \sf{ \frac{ {a}^{2} }{ \sqrt{ {a}^{2} + {b}^{2} } + b} = \sqrt{ {a}^{2} + {b}^{2} } - b}$

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Learn more from Brainly :-

1. If x=√3a+2b + √3a-2b / √3a+2b - √3a-2b

prove that bx²-3ax+b=0

2. if the real POSITIVE FOURTH root of(28+16√3) is divided by (2+√3) and the result is EXPRESSED as (a+b√3) with a&b both

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