plz answer this question plz step
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rationalize all the terms
as
=> 6/(2√3-√6)
= 6(2√3+√6)/(12-6)
= 2√3+√6
=> √6/(√3+√2)
= √6(√3-√2)
= 3√2-2√3
=> 4√3/(√6-√2)
= 4√3(√6+√2)/4
= 3√2 + √6
so
(2√3+√6)+(3√2-2√3)-(3√2+√6)
= 0
hope it helps you
@di
as
=> 6/(2√3-√6)
= 6(2√3+√6)/(12-6)
= 2√3+√6
=> √6/(√3+√2)
= √6(√3-√2)
= 3√2-2√3
=> 4√3/(√6-√2)
= 4√3(√6+√2)/4
= 3√2 + √6
so
(2√3+√6)+(3√2-2√3)-(3√2+√6)
= 0
hope it helps you
@di
Answered by
1
Here.... Hope it helps
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