Find the value of m and n: if:
5+237+43=m+n3\dfrac{5 + 2\sqrt3}{7 + 4\sqrt3} = m + n\sqrt37+435+23=m+n3
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5+237+43=5+237+43×7−437−43=(5+23)×(7−43)(7+43)×(7−43)=5×(7−43)+23×(7−43)(7)2−(43)2=35−203+143−2449−48=11−631=11−63\dfrac{5 + 2\sqrt3}{7 + 4\sqrt3} = \dfrac{5 + 2\sqrt3}{7 + 4\sqrt3} \times \dfrac{7 - 4\sqrt3}{7 - 4\sqrt3} \\[1em] = \dfrac{(5 + 2\sqrt3) \times (7 - 4\sqrt3)}{(7 + 4\sqrt3) \times (7 - 4\sqrt3)}\\[1em] = \dfrac{5\times(7 - 4\sqrt3) + 2\sqrt3 \times (7 - 4\sqrt3)}{(7)^2 - (4\sqrt3)^2}\\[1em] = \dfrac{35 - 20\sqrt3 + 14\sqrt3 - 24}{49 - 48}\\[1em] = \dfrac{11 - 6\sqrt3}{1}\\[1em] = 11 - 6\sqrt3\\[1em]7+435+23=7+435+23×7−437−43=(7+43)×(7−43)(5+23)×(7−43)=(7)2−(43)25×(7−43)+23×(7−43)=49−4835−203+143−24=111−63=11−63
Given :5+237+43=m+n3\dfrac{5 + 2\sqrt3}{7 + 4\sqrt3} = m + n\sqrt37+435+23=m+n3
⇒ 11−63=m+n311 - 6\sqrt3 = m + n \sqrt311−63=m+n3
Hence, m = 11 and n = -6.
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Simplify :
717−23−317+23\dfrac{7}{\sqrt{17} - 2\sqrt{3}} - \dfrac{3}{\sqrt{17} + 2\sqrt{3}}17−237−17+233
3+23−2=m+n2\dfrac{3 + \sqrt2}{3 - \sqrt2} = m + n \sqrt23−23+2=m+n2
By rationalising the denominator of each of the following; find, in each case, the value correct to two significant figures :
(i) 13−2\dfrac{1}{3 - \sqrt2}3−21
(ii) 12+3\dfrac{1}{2 + \sqrt3}2+31
(iii) 432−23\dfrac{4}{3\sqrt2 - 2\sqrt3}32−234
Calculate the compound interest on ₹ 18,000 at 10% per annum in two years.
