Question

Add:
6m²n +4mn
an² +5,
n²- nm² +3,
mn
3n² – amin-4.​

Answer 1

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

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