$\frac12\left\{\left(a+b\right)^2+\left(a-b\right)^2\right\}$ = কত?
Options :
- (A) $a^2+b^2$
- (B) $\frac{\left(a+b\right)^2}2-\frac{\left(a-b\right)^2}2$
- (C) $a^2-b^2$
- (D) $\left(a+b\right)^2+\left(a-b\right)^2$
1 টি উত্তর পাওয়া গেছে
$\frac12\left\{\left(a+b\right)^2+\left(a-b\right)^2\right\}$
= $\frac12\left\{a^2+2ab+b^2+a^2-2ab+b^2\right\}$
= $\frac12\left\{a^2+\cancel{2ab}+b^2+a^2-\cancel{2ab}+b^2\right\}$
= $\frac12\left\{2a^2+2b^2\right\}$
= $\frac12.2\left\{a^2+b^2\right\}$
= $a^2+b^2$
= $\frac12\left\{a^2+2ab+b^2+a^2-2ab+b^2\right\}$
= $\frac12\left\{a^2+\cancel{2ab}+b^2+a^2-\cancel{2ab}+b^2\right\}$
= $\frac12\left\{2a^2+2b^2\right\}$
= $\frac12.2\left\{a^2+b^2\right\}$
= $a^2+b^2$