Given (x + my) (x + NY) = x ^ 2-5xy + 3x ^ 2, find the algebraic formula 3mn-2 (M + n)

Given (x + my) (x + NY) = x ^ 2-5xy + 3x ^ 2, find the algebraic formula 3mn-2 (M + n)

Wrong title: it should be (x + my) (x + NY) = x ^ 2-5xy + 3Y ^ 2
(x+my)(x+ny)=x^2-5xy+3y^2
x²+（m+n）xy+mny²=x²-5xy+3y²
m+n=-5
mn=3
So: 3mn-2 (M + n) = 3 × 3-2 × (- 5) = 9 + 10 = 19

If the sum of solutions of 3x + 5Y = 2m + 2, x + 3Y = m is equal to 2, then m ^ 2-2m

y=m/4-0.5
x=m/4+1.5
So m / 2 + 1 = 2, M = 2
So m ^ 2-2m = 0

The value range of integer a is 2

Radical a + 1 and radical 2 are the same kind of quadratic radical
So a + 1 = 2x & # 178;
a=2x²-1
So x = 2, a = 7
x=3,a=17

If a ^ 2-6a + 9 = 3 under a + radical, then the value range of a is___

a+√(a-3)²=3
a+|a-3|=3
Then a + 3-A = 3
So | A-3 | = 3-A
So A-3 ≤ 0
a≤3