1.

II. y3 – \(\frac{{\left( {3\;\times \;8} \right)\frac{9}{2}}}{{y\sqrt y }}\) = 01). if x > y2). if x ≥ y3). if x < y4). if x ≤ y

Answer»

I. $(\frac{{17}}{{\sqrt X }} = \sqrt x - \frac{7}{{\sqrt x }} = 0)$

⇒ $(\frac{{17}}{{\sqrt x }})$ + $(\frac{7}{{\sqrt x }} = \sqrt x)$

⇒ $(\frac{{17\; + \;7}}{{\sqrt x }})$ = √x

24 = x

II. y3 – $(\frac{{\left( {3\; \times \;8} \right)\frac{9}{2}}}{{y\sqrt y }})$ = 0

⇒ y9/2 = 249/2

⇒ y = 24

When x = 24, x = y for y = 24

∴ x = y


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