Question

plz don't give silly answers ​

Answer 1

Question:-

Prove that:

$\sf \large \bigg( \frac{4}{3} m - \frac{3}{4}n \bigg) {}^{2} + 2mn = \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} .$

Given:-

$\sf \large \bigg( \frac{4}{3} m - \frac{3}{4}n \bigg) {}^{2} + 2mn = \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} .$

To Prove:-

$\sf \large \bigg( \frac{4}{3} m - \frac{3}{4}n \bigg) {}^{2} + 2mn = \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} .$

Solution:-

L.H.S.

$\sf \large = \bigg( \frac{4}{3} m \bigg) {}^{2} + \bigg( \frac{3}{4} n \bigg) {}^{2} - 2 \times \frac{4}{3} m \times \frac{3}{4} n + 2mn$

$\sf \large = \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} - 2 \times \frac{4}{3} \times \frac{3}{4} mn + 2mn$

$\sf \large = \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} - 2mn + 2mn$

$\sf \large = \bigg( \frac{16}{9} m {}^{2} + \frac{9}{16} n {}^{2} \bigg)R.H.S.$

Answer:-

Hence, Proved that L.H.S. = R.H.S.

Hope you have satisfied. ⚘