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प्रश्न
Show that: \[\left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2mn = \frac{16 m^2}{9} + \frac{9 n^2}{16}\]
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उत्तर
\[\text { LHS } = \left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2mn\]
\[ = \left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2 \times \frac{4m}{3} \times \frac{3n}{4}\]
\[ = \left( \frac{4m}{3} \right)^2 + \left( \frac{3n}{4} \right)^2 \left[ \because \left( a - b \right)^2 + 2ab = a^2 + b^2 \right]\]
\[ = \frac{16 m^2}{9} + \frac{9 n^2}{16}\]
= RHS
Because LHS is equal to RHS, the given equation is verified.
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