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Rationalise 1/3-root7+root3 |
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Answer» -\frac{1}{20}(2\sqrt{7}-6\sqrt{3}+8\sqrt{2}-2\sqrt{42}) Step-by-step explanation: Here, the given EXPRESSION is, \frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} } Multiply and divide by √7 + √3 + √2, =\frac{1}{\sqrt{7} +\sqrt{3} -\sqrt{2} }\times \frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{\sqrt{7} +\sqrt{3} +\sqrt{2}} =\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{(\sqrt{7} +\sqrt{3})^2-(\sqrt{2})^2 } =\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{7+3+2\sqrt{21}-2} =\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}} Multiply and divide by 8-2√21, =\frac{\sqrt{7} +\sqrt{3} +\sqrt{2}}{8+2\sqrt{21}}\times \frac{8-2\sqrt{21}}{8-2\sqrt{21}} =\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-2\sqrt{147} -2\sqrt{63} -2\sqrt{42}}{(8)^2-(2\sqrt{21})^2} =\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{64-84} =\frac{8\sqrt{7} +8\sqrt{3} +9\sqrt{2}-14\sqrt{3} -6\sqrt{7} -2\sqrt{42}}{-20} =\frac{2\sqrt{7} -6\sqrt{3} +9\sqrt{2}-2\sqrt{42}}{-20} =-\frac{1}{20}(2\sqrt{7}-6\sqrt{3}+8\sqrt{2}-2\sqrt{42}) |
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