Question

(6x+12)/(x^2+4x+4)+(4-x^2)/(x^2-4x+4)=(x^2)/(4-x^2)

Answer 1

化简左式：

(6x+12)/(x^2+4x+4)+(4-x^2)/(x^2-4x+4)

=6(x+2)/(x+2)²+(2+x)(2-x)/(x-2)²

=6/(x+2)-(x+2)/(x-2)

=6(x-2)/[(x+2)(x-2)]-(x+2)²/[(x+2)(x-2)]

=6(x+2)/(x²-4)-(x+2)²/(x²-4)

=(2x-x²-16)/(x²-4)

=(x²-2x+16)/(4-x²)

∵左式=右式

∴(x²-2x+16)/(4-x²)=x²/(4-x²)=

∴x²-2x+16=x²

∴x=8