Question

Add :
6m^2n + 4mn – 2n^2 + 5, n^2 – nm^2 + 3, mn – 3n^2 – 2m^2n – 4

Answer 1

$\textbf{Answer :}$

The three terms are

(6m²n + 4mn - 2n² + 5),
(n² - nm² + 3) and
(mn - 3n² - 2m²n - 4)

So, the required sum be

= (6m²n + 4mn - 2n² + 5)
+ (n² - nm² + 3)
+ (mn - 3n² - 2m²n - 4)

= (6 - 2)m²n + (4 + 1)mn
+ (- 2 + 1 - 3)n² - nm²
+ (5 + 3 - 4)

= 4m²n + 5mn - 4n² - nm² + 4

#$\textbf{MarkAsBrainliest}$

Answer 2

Hiii. .....friend

The answer is here,

Addition :-

$= > \: 6 {m}^{2} n + 4mn - 2 {n}^{2} + 5 + {n}^{2} - n {m}^{2} + 3 + mn - 3 {n}^{2} - 3 {m}^{2} n - 4$

$= > \: (6 - 1 - 2) {m}^{2} n + (1 - 2 - 3) {n}^{2} + (4 + 1)mn + 5 - 4+3$

$= > \: 3 {m}^{2} n - 4 {n}^{2} + 5mn + 4$

:-)Hope it helps u.