數學 - 恆等式

爆開2邊

LHS
m(x^2 + 9 - 6x) + nx + 1
= (m)x^2 + (n - 6m)x+ (9m+1)
比較2邊
m = 1
n - 6m = -38
9m + 1 = 41

之後就發現
你係唔係打錯題目= =
m=1 => 9m+1=10 // n - 6 = -38 => n= -32
答案太奇怪==...

m(x - 3)^2 + n(x + 1) ^2
=m(x^2 + 9 - 6x) + n( x^2 + 1 +2x)
=(m+n)x^2 + (2n-6m)x + (9m+n)

比較左右2邊
x^2既coeff = x^2既coeff
x既coeff = x既coeff
constant = constant

所以就有
m+n = 1
2n-6m = -38
9m+n = 41
是旦解2條

之後我地就會有 (解1同2)
n=-4 m=5
咁9m+n 亦=41

所以n就係=-4
m=5了

m(x - 3)^2 + n(x + 1)^2 ≡ x^2 - 38x + 41
左=m(x^2+6x+9)+n(x^2+2x+1)
左=mx^2+6mx+9m+nx^2+2nx+n
左=(m+n)x^2+x(6m+2n)+(9m+n)
m+n=1
9m+n=41
ii-i
8m=40
m=5

The only x^2 term on LHS is mx^2.
The only x^2 term on RHS is x^2
so m = 1.