The value of \(\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}\) is = ?
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1]
0
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1
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\({\left[ {2\left( {x + y + z} \right)} \right]^{ - 1}}\)
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\({\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}}\)
Solution
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98643 – 21748 = 51212 + ?
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Find the value of \(\frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}} + \left[ {0.75 + {{\left( {0.75} \right)}^2} + 1} \right]\)
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5
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Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
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The value of x in the equation
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\(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{{14}} + \frac{1}{{28}}\) is equal to = ?
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If \(\sqrt {{ \text{4096}}}\) = 64, then the value of \(\sqrt {{ \text{40}}{ \text{.96}}}\) + \(\sqrt {{ \text{0}}{ \text{.4096}}}\) + \(\sqrt {{ \text{0}}{ \text{.004096}}}\) + \(\sqrt {{ \text{0}}{ \text{.00004096}}}\) up to two place of decimals is = ?
Solution |
| # | Quiz |
= 5