If Mn is not equal to 0, then the sum of | m | / M and N / | n | cannot be a.0 B.1 C.2 D. - 2 Better solve it today! | Math enthusiast | Mathematical art

If Mn is not equal to 0, then the sum of | m | / M and N / | n | cannot be a.0 B.1 C.2 D. - 2 Better solve it today!

B
the reason being that
From m * n ≠ 0, we get m ≠ 0, n ≠ 0
So | m | / M can only be equal to 1 or - 1, and so can n / | n |
So the value of | m | / M + n / | n | is only - 2, 0 and 2

If n of M-N = a, when Mn is 5 times larger, n of M-N is 1=

5n/5m-5n=a,
All 5 in the above formula can be omitted,
And N / M-N = a,
O(∩_ ∩) O, hope to help you, hope to adopt

If (Mn + m-n-n-n / n-1) * (m-1 / M-N + m-mn-n) * n-1 / M-1 is a fixed value, request the fixed value

The fixed value is 1
Because:
(mn+m-n^2-n)/(n^2-1)=[(n+1)*(m-n)]/[(n+1)*(n-1)]=(m-n)/(n-1)；
(m^2-1)/(m^2+m-mn-n)=[(m+1)*(m-1)]/[(m+1)*(m-n)]=(m-1)/(m-n).
So:
[(mn+m-n^2-n)/(n^2-1)]*[(m^2-1)/(m^2+m-mn-n)]*[(n-1)/(m-1)]
=[(m-n)/(n-1)]*[(m-1)/(m-n)]*[(n-1)/(m-1)]=[(m-n)*(m-1)*(n-1)]/[(n-1)*(m-n)*(m-1)]=1

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