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Rationalise the denominator of `1/(sqrt3-sqrt2-1)`

Answer» `1/(sqrt3-sqrt2-1)=1/(sqrt3-sqrt2-1)xx(sqrt3-sqrt2+1)/(sqrt3-sqrt2+1)=(sqrt3-sqrt2+1)/((sqrt3-sqrt2-1)(sqrt3-sqrt2-1))`
`(sqrt3-sqrt2+1)/((sqrt3-sqrt2)^(2)-1^(2))`
`=(sqrt3-sqrt2+1)/(3+2-2sqrt3xxsqrt2-1)=(sqrt3-sqrt2+1)/(4-2sqrt6)`
`=(sqrt3-sqrt2+1)/(2(2-sqrt6))=(sqrt3-sqrt2+1)/(2(2-sqrt6))xx(2+sqrt6)/(2+sqrt6)`
`((sqrt3-sqrt2+1)(2+sqrt6))/(2(2^(2)-(sqrt6)^(2)))`
`(2sqrt3-2sqrt2+2+sqrt3xxsqrt6-sqrt2xxsqrt6+sqrt6)/(2(4-6))`
`=(2sqrt3-2sqrt2+2+sqrt(3xx6)-sqrt(2xx6)+sqrt6)/-4`
`=(2sqrt3-2sqrt2+2+sqrt(3^(2)xx2)-sqrt(2^(2)xx3)+sqrt6)/-4`
`=(2sqrt3-2sqrt2+2+3sqrt2-2sqrt3+sqrt6)/-4`
`=(sqrt2+sqrt6+2)-4=-((sqrt2+sqrt6+2)/4)`


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