If (M + n) ^ 2-MN (M + n) = (M + n) · m, then M is () a.m ^ 2 + n ^ 2 B.M ^ 2-MN + n ^ 2 C.M ^ 2-3mn + n ^ 2 D.M ^ 2 + Mn + n ^ 2

If (M + n) ^ 2-MN (M + n) = (M + n) · m, then M is () a.m ^ 2 + n ^ 2 B.M ^ 2-MN + n ^ 2 C.M ^ 2-3mn + n ^ 2 D.M ^ 2 + Mn + n ^ 2

(m+n)^2-mn(m+n)=（m+n)(m+n-mn)=(m+n)·M
[none of the answers above]
(m+n)^3-mn(m+n)=（m+n)(m^2+n^2+mn)=(m+n)·M
D.m^2+mn+n^2

Let f (x) = loga (1-x) / (1 + x), a > 0, a ≠ 1. For any m, n ∈ (- 1,1) f (m) + F (n) = f [(M + n) / (1 + Mn)]

f(m) + f(n) = loga(1-m)/(1+m) + loga(1-n)/(1+n)
= loga[(1-m)(1-n)]/[(1+m)(1+n)]
=loga[1+mn - m - n]/[1+mn+m+n]
=loga[ 1 - (m+n)/(1+mn)]/[1 + (m+n)/(1+mn)]
= f[(m+n)/(1+mn)]
The proof is complete

Given that x = 3, y = - 2 is the solution of the equations ax + by = 6, 2ax-3by = 12, find the value of a-2b

Substituting x = 3 and y = - 2 into the equation, we can get the equation of a and B, that is:
3a-2b = 6 and 6A + 6B = 12 (a + B = 2),
The solution is a = 2, B = 0,
So a-2b = 2

Solving the equations ax + by = ab 2ax-3by = 12ab

ax+by=ab （1）
2ax-3by=12ab（2）
(1) * 2 - (2) de
5by=-10ab
y=-2a
(1) * 3 + (2) gain
5ax=15ab
x=3b