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Question 52 x...

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Question 52

$x^{m} \times y^{m} = (x \times y)^{2 m},$ where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.

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Solution

False
If $a$ and $b$ are rational numebrs, then
$a^{m} \times b^{m} = (a b)^{m} ∴ x^{m} \times y^{m} = (x y)^{m} = (x \times y)^{m} H e n c e, x^{m} \times y^{m} \neq (x \times y)^{2 m}$

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Question 51

$x^{m} + x^{m} = x^{2 m},$ where $x$ is a non-zero rational number and $m$ is a positive integer.

x^{m} \times y^{m} = (x y)^{2 m}

x^{m} \div y^{m} = {(\frac{x}{y})}^{1}

3.8 \times 10^{7}

(- 16 \times 2)^{4}

6^{2}

2^{6}

\frac{x^{n} y^{m}}{(1 + x^{2 n}) (1 + y^{2 m})} > \frac{1}{4}

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