rationalise 1/1+√2-√3
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[1/1+√2-√3][1+√2+√3/1+√2+√3]
=1+√2+√3/(1+√2)^2-3
=1+√2+√3/1+2+2√2-3
=1+√2+√3/2√2
=[1+√2+√3/2√2][√2/√2]
=√2+4+√6/4
=4+√2(1+√3)/4
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