Question

Answer these questions 23,24,25 with brief explanation...

Answer 1

Hey buddy! Here is the solutions for the questions you asked.

$( 1) \\ \frac{ {x}^{ 1 + \frac{1}{2} } + xy }{xy - {y}^{3} } - \frac{ \sqrt{x} }{ \sqrt{x} + y } \\ \frac{(x \sqrt{x} + xy) ( \sqrt{x} - y) - \sqrt{x} (xy - {y}^{3}) }{(xy - {y}^{3})( \sqrt{x} - y ) } \\ \frac{ {x}^{2} + xy \sqrt{x} - xy \sqrt{x} - x {y}^{2} - xy \sqrt{x} + {y}^{3} \sqrt{x} }{xy \sqrt{x} - {y}^{3} \sqrt{x} - x {y}^{2} + {y}^{4} } \\ \frac{x(x - {y}^{2}) - y \sqrt{x} (x - {y}^{2} )}{y \sqrt{x}(x - {y}^{2}) - {y}^{2}(x - {y}^{2} ) } \\ \frac{(x - y \sqrt{x} )(x - {y}^{2} )}{(y \sqrt{x} - {y}^{2} )(x - {y}^{2}) } \\ \frac{ \sqrt{x} ( \sqrt{x} - y)}{y( \sqrt{x} - y) } \\ \frac{ \sqrt{x} }{y} (answer) \\ \\ \\ (2 ) \\ \\ {(c)}^{z} = a \\ ( {{b}^{y}) }^{z} = a \\ {(b)}^{yz} = a \\ { ({a}^{x} )}^{yz} = a \\ {a}^{xyz} = a \\ xyz = 1 (answer)\\ \\ \\ (3) \\ \frac{ {a}^{2} + \frac{1}{ {a}^{2} } + 2 }{ {a}^{3} + \frac{1}{ {a}^{3} } + 3a + \frac{3}{a} } \\ \frac{( {a + \frac{1}{a} })^{2} }{ {(a + \frac{1}{a}) }^{3} } \\ {(a + \frac{1}{a} )}^{ - 1} \\ {( \frac{ {a}^{2} + 1)}{a} }^{ - 1} \\ \frac{a}{ {a}^{2} + 1 } (answer)$
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Gaurav kumar