Simplify : \(1 + {2 \over {1 + {3 \over {1 + {4 \over 5}}}}}\)
7/4
4/7
7/5
3/7
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1
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Simplify : \(\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right] \div \left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right] = ?\)
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2
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Let 0 < x < 1, then the correct inequality is = ?
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3
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The value of \(\frac{5}{{1\frac{7}{8}{ \text{of 1}}\frac{1}{3}}} \times \frac{{2\frac{1}{{10}}}}{{3\frac{1}{2}}}{ \text{ of 1}}\frac{1}{4} = ?\)
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4
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The expression \(\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}\) is equal to = ?
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5
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\({ \text{If }}x = \frac{1}{{2 + \frac{1}{2}}}{ \text{ then }}\frac{1}{x} = ?\)
Solution |
6
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If a - b = 3 and a2 + b2 = 29, find the value of ab = ?
Solution |
7
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A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
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8
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(36.14)^2 – (21.28)^2 =?
Solution |
9
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Simplify : \(\frac{{\frac{5}{3} \times \frac{7}{{51}}{ \text{ of }}\frac{17}{5} - \frac{1}{3}}}{{\frac{2}{9} \times \frac{5}{7}{ \text{ of }}\frac{{28}}{5} - \frac{2}{3}}}{\kern 1pt} {\kern 1pt} = ?\)
Solution |
10
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5907 – 1296 / 144 = x * 8.0
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