Given the linear function y = (2-m) x-2m & sup2; + 8, when the value of M is, the image passes through the point (- 1,5)

Given the linear function y = (2-m) x-2m & sup2; + 8, when the value of M is, the image passes through the point (- 1,5)

If y = (2-m) x-2m2 + 8, m-2-2m2 + 8 = 5
The solution is: M = 1 or - 1 / 2

When 3M = 2n, then (M / M + n) + (n / m-n) - (N2 / m2-n2) is equal to?

3m=2nm=2n/3m/(m+n)+n/(m-n)-(n^2)/(m^2-n^2)=[m(m-n)+n(m+n)]/(m^2-n^2)-(n^2)/(m^2-n^2)=(m^2+n^2)/(m^2-n^2)-(n^2)/(m^2-n^2)=m^2/(m^2-n^2)=[4(n^2)/9]/[4(n^2)/9-n^2]=-4/5

Let a = m 2 + N 2, B = x 2 + y 2, a and B be normal numbers, and a is not equal to B, find the maximum value of m x + y 2

By using the mean inequality, we can get the following results
Because MX