a/(x+2)与b/(x-2)的和为4x/(x的2次幂-4),a=?,b=?

a/(x+2)与b/(x-2)的和为4x/(x的2次幂-4),a=?,b=?
a/(x+2)与b/(x-2)的和为4x/(x的2次幂-4),a=?,b=?

a/(x+2)与b/(x-2)的和为4x/(x的2次幂-4),a=?,b=?
a/(x+2)+b/(x-2)=4x/(x*x-4)
[a(x-2)+b(x+2)]/(x*x-4)=4x/(x*x-4)
[(a+b)x-(2a-2b)]/(x*x-4)=4x/(x*x-4)
要使对于一切有意义的x等式恒成立,就使
a+b=4
2a-2b=0
a=2
b=2

通分得
a(x-2)+b(x+2)/(x+2)(x-2)=4x/(x+2)(x-2)
去括号
ax-2a+bx+2b/(x+2)(x-2)=4x/(x+2)(x-2)
因为分母相同所以
ax-2a+bx+2b=4x
x(a+b)-2(a-b)=4x
所以得到2元一次方程
a+b=4
a-b=0
解得a=2 b=2