m(x+2)/(x+1)(x+2)-n(x-1)/(x+1)(x+2)=1/x²+3x+2 m、n不为0的常数,求1/(m+n)²
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m(x+2)/(x+1)(x+2)-n(x-1)/(x+1)(x+2)=1/x²+3x+2 m、n不为0的常数,求1/(m+n)²
m(x+2)/(x+1)(x+2)-n(x-1)/(x+1)(x+2)=1/x²+3x+2 m、n不为0的常数,求1/(m+n)²
m(x+2)/(x+1)(x+2)-n(x-1)/(x+1)(x+2)=1/x²+3x+2 m、n不为0的常数,求1/(m+n)²
(x+1)(x+2)=x^2+3x+2
两边分母相同,比较分子的系数即可
左边分子为
mx+2m-(nx-n)
=(m-n)x+(2m+n)
右边分子为1
所以
m-n=0
2m+n=1
解出m=1/3,n=1/3
1/(m+n)^2=9/4
如仍有疑问,欢迎追问!
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