Question

दर्शाइए कि:

${\displaystyle {(\frac{4}{3}\text{m}-\frac{3}{4}\text{n})}^2+2\text{mn}=\frac{16}{9}{\text{m}}^2+\frac{9}{16}{\text{n}}^2}$

Answer 1

L.H.S.= ${\displaystyle {(\frac{4}{3}\text{m}-\frac{3}{4}\text{n})}^2+2\text{mn}}$

सर्वसमिका (a−b)² = a² − 2ab + b² से

${\displaystyle {\left(\frac{4}{3}\text{m}\right)}^2-2\times \frac{4}{3}\text{m}\times \frac{3}{4}\text{n}+{\left(\frac{3}{4}\text{n}\right)}^2+2\text{mn})}$

=${\displaystyle \frac{16}{9}{\text{m}}^2-2\text{mn}+\frac{9}{16}{\text{n}}^2+2\text{mn}}$

=${\displaystyle \frac{16}{9}{\text{m}}^2+\frac{9}{16}{\text{n}}^2}$

= R.H.S. 

अत: L.H.S.= R.H.S.