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Rationalise the denominator of:
(3√2 - 2√3)/(√8 + √27)​


Rationalise The Denominator Of32 238 27 class=

Sagot :

Step-by-step explanation:

[tex] \large{ \sf \underline{Solution - }}[/tex]

[tex] \sf To \: rationalise \: = \: \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{\sqrt{8} + \sqrt{27} } [/tex]

[tex] \sf \dashrightarrow \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{2 \sqrt{2} + 3 \sqrt{3} } [/tex]

[tex] \sf \dashrightarrow \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{2 \sqrt{2} + 3 \sqrt{3} } \times \dfrac{2 \sqrt{2} - 3 \sqrt{3}}{2 \sqrt{2} - 3 \sqrt{3}} [/tex]

[tex] \sf Combine \: the \: fractions, [/tex]

[tex] \sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{(2 \sqrt{2} + 3 \sqrt{3} )(2 \sqrt{2} - 3 \sqrt{3})} [/tex]

[tex] \sf \underline{Solving \: denominator} , [/tex]

[tex] \sf We \: know \: that, [/tex]

[tex]\sf \dashrightarrow \boxed{ \tt (a + b)(a - b) = a^{2} - b ^{2}}[/tex]

[tex] \bf So , [/tex]

[tex] \sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{(2 \sqrt{2})^{2} - (3 \sqrt{3} )^{2} } [/tex]

[tex] \sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{8 - 27 } [/tex]

[tex] \sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{ - 19 } [/tex]

[tex] \sf \underline{Solving \: numerator}, [/tex]

[tex] \sf \dashrightarrow \dfrac{3 \sqrt{2} (2 \sqrt{2} - 3 \sqrt{3})- 2 \sqrt{3}(2 \sqrt{2} - 3 \sqrt{3})}{ - 19 } [/tex]

[tex] \sf \dashrightarrow \dfrac{12 - 9 \sqrt{6} - 4 \sqrt{6} + 18}{ - 19 } [/tex]

[tex] \sf \dashrightarrow \dfrac{30 - 13 \sqrt{6} }{ - 19 } [/tex]

[tex] \sf \dashrightarrow - \dfrac{30 - 13 \sqrt{6} }{ 19 } [/tex]

[tex] \bf Hence, [/tex]

[tex] \sf On \: rationalising \: we \: got,[/tex]

[tex] \bf \dashrightarrow - \dfrac{30 - 13 \sqrt{6} }{ 19 }[/tex]