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\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + n√6 \(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\) \(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\) \(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\) = \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\) = \(\frac{0 - 4 \sqrt{6} - 6}{5}\) = \(\frac{10 - 5 \sqrt{6}}{5}\) = − 2 + √6 ∴ m + n\(\sqrt{6}\) = − 2 + √6 m = − 2, n = 1

\(\frac{\log 7 \times \log 10}{\log 2}\) log 7 x log 10 \(\div\) log 2 a + 1 - b a - b + 1

\(\sqrt{(x_2 - x_1)^2 + (y_2 - Y_1)^2}\) = \(\sqrt{(2 - 2)^2 + (5 - 3)^2}\) = \(\sqrt{0^2 + 2^2}\) = \(\sqrt{4}\) = 2 units

\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\) \(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\) 2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4 = 4

t = \(2^{-x} = \frac{1}{2^{x}}\) \(\implies 2^{x} =\frac{1}{t}\) \(2^{x+1} = 2^{x} \times 2^{1}\) = \(\frac{1}{t} \times 2 = \frac{2}{t}\)

\(60^2 + x^2 + x^2\) \(360^2 = 2x^2\) \(x^2\) = 1800 x = \(\sqrt{1800}\) x = 42.4264 x = 42.4264 perimeter = 4x = 4 x 42.4264 = 169.7056 = 120\(\sqrt{2}\) = 120\(\sqrt{2}\)

(a) \((y - 1)\log_{10} 4 = y\log_{10} 16\) \((y - 1)\log_{10} 4 = y \log_{10} 4^{2}\) \((y - 1)\log_{10} 4 = 2y\log_{10} 4\) Equating both sides, we have \(y - 1 = 2y \implies -1 = 2y - y\) \(\therefore y = -1\) (b) Let the distance from my house to the office = c. At 4km/h, the time taken to get to the office from the house = \(\frac{c}{4} hr\) At 5km/h, the time taken to get to the office from the house = \(\frac{c}{5} hr\) \(\frac{c}{4} = \frac{c}{5} + \frac{30}{60}\) \(\frac{c}{4} - \frac{c}{5} = \frac{1}{2}\) \(\frac{c}{20} = \frac{1}{2} \implies c = 10km\)

\(\frac{0.026 \times 0.36}{0.69}\) = \(\frac{26 \times 10^-3 \times 36 \times 10^-2}{69 \times 10^-2}\) = 0.01356 = 1.36 x 10-2

\(\frac{27^\frac{1}{3} - 8^\frac{2}{3}}{16^\frac{2}{4} \times 2}\) \(\frac{27^\frac{1}{3} - 8^\frac{2}{3}}{16^\frac{2}{4} \times 2}\) = \(\frac{3\sqrt{27} - 3\sqrt{8}^2}{\sqrt{16} \times 2}\) \(\frac{3 - 4}{4 \times 2}\) = \(\frac{-1}{8}\)

\(\frac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}}\) = \(\frac{6-2\sqrt{6} - 2\sqrt{6} + 4}{6 - 4}\) \(\frac{10 - 4\sqrt{6}}{2}\) = 5 - 2\(\sqrt{6}\)

Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\) = \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\) hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\)) = 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\) = 4 x 15 - 2 x 15 - 2 x 15 + 15 = 60 - 30 - 30 + 15 = 15

\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\) = \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n - 1 + 3n + 3 - 4n}\) = 3\(^{4n - 4n - 1 + 3}\) = 32 = 9

|PQ| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\) = \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\) = \(\sqrt{4^2 + 6^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = 2\(\sqrt{13}\)

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\) \(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\)) = 30o In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1 \(\beta\) = tan-1(1) = 45o Hence, o + 45o = 75o

If logx 64 = 3, then \(64 = x^3\) \(4^3 = x^3\) Since the indices are equal, x = 4 Hence, x log\(_2\)8 = 4(3.log\(_2\)2) = where log\(_2\)2 = 1(1) = 12 X 1 = 12