Question

After rationalising the denominator of 22÷2+√3+√5 we get?​

Answer 1

Answer:

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Step-by-step explanation:

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