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प्रश्न
Simplify: \[\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3} - \sqrt{2}}\]
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उत्तर
We know that rationalization factor for `3sqrt2 +2sqrt3` and `sqrt3-sqrt2`are `3sqrt2 - 2sqrt3`and `sqrt3 +sqrt2`respectively. We will multiply numerator and denominator of the given expression `(3sqrt2 -2sqrt3)/ (3sqrt2 +2sqrt3)` and `sqrt12/(sqrt3-sqrt2)`by `3sqrt2-2sqrt3`and `sqrt3 +sqrt2`respectively, to get
`(3sqrt2 -2sqrt3)/ (3sqrt2 +2sqrt3) xx (3sqrt2 -2sqrt3)/ (3sqrt2 -2sqrt3) +sqrt12/(sqrt3-sqrt2) xx (sqrt3 +sqrt2)/(sqrt3 +sqrt2) =((3sqrt2)^2+(2sqrt3)^2 - 2xx 3 sqrt2 xx 2sqrt3)/((3sqrt2)^2 -( 2sqrt3)^2 ) +(sqrt36+sqrt24)/((sqrt3)^2-(sqrt2)^2)`
`= (18+12-12sqrt6)/(18-12) +(6+sqrt24)/(3-2)`
`= (30-12sqrt6 +36 +12sqrt6)/6`
`=66/6`
`=11`
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संबंधित प्रश्न
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = `c/d`. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
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If x= \[\sqrt{2} - 1\], then write the value of \[\frac{1}{x} . \]
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