rationalising the denominator of 22/2+√3+√5 step by step
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Answered by
64
Hi,
22/( 2 + √3 + √5 )
= [22(2+√3-√5)]/[(2+√3+√5)(2+√3-√5)]
= [22(2+√3-√5)]/[(2+√3)²-(√5)²]
=[22(2+√3-√5)]/[4+3+4√3-5]
=[22(2+√3-√5)]/[2+4√3]
=[11(2+√3-√5)]/[1+2√3]
=[11(2+√3-√5)(2√3-1)]/[(2√3+1)(2√3-1)]
=[11(4√3-2+6-√3-2√15+√5]/(12-1)
= [11(4+3√3-2√15)]/11
= 4 + 3√3 - 2√15
I hope this helps you.
: )
22/( 2 + √3 + √5 )
= [22(2+√3-√5)]/[(2+√3+√5)(2+√3-√5)]
= [22(2+√3-√5)]/[(2+√3)²-(√5)²]
=[22(2+√3-√5)]/[4+3+4√3-5]
=[22(2+√3-√5)]/[2+4√3]
=[11(2+√3-√5)]/[1+2√3]
=[11(2+√3-√5)(2√3-1)]/[(2√3+1)(2√3-1)]
=[11(4√3-2+6-√3-2√15+√5]/(12-1)
= [11(4+3√3-2√15)]/11
= 4 + 3√3 - 2√15
I hope this helps you.
: )
Answered by
32
HERE is ur answer
4+3√3+√5-2√15
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