Rationalise the denominator of 1\underroot 3-under root 2
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Answered by
3
hii :)
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here
1 = 1 × √3+√2
√3-√2 √3-√2 √3+√2
=√3+√2 (using a²-b²= (a+b)(a-B)
3-2
√3+√2
1
____________________________________________________________
here
1 = 1 × √3+√2
√3-√2 √3-√2 √3+√2
=√3+√2 (using a²-b²= (a+b)(a-B)
3-2
√3+√2
1
Answered by
2
Rationalising 1/(✓3-✓2) →
1/(✓3-✓2) × (✓3+✓2)/(✓3+✓2)
= (✓3+✓2)/(✓3)²-(✓2)²
( by identity → (a-b)(a+b) = a²-b²)
= (✓3+✓2)/(3-2)
= (✓3+✓2)/1
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1/(✓3-✓2) × (✓3+✓2)/(✓3+✓2)
= (✓3+✓2)/(✓3)²-(✓2)²
( by identity → (a-b)(a+b) = a²-b²)
= (✓3+✓2)/(3-2)
= (✓3+✓2)/1
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