1.已知1/x-1/y=3 求代数式2x-14xy-2y/x-2xy-y的值2.分解因式(x+2)(x+4)+x²-4
问题描述：
1.已知1/x-1/y=3 求代数式2x-14xy-2y/x-2xy-y的值
2.分解因式(x+2)(x+4)+x²-4
答
1.已知1/x-1/y=3 2x-14xy-2y/x-2xy-y=[2(x-2xy-y)-10xy]/(x-2xy-y)=2- 10xy/(x-2xy-y)=2+10xy/5xy=2+2=42.分解因式(x+2)(x+4)+x²-4=（x+2)(x+4)+(x-2)(x+2)=(x+2)(x+4+x-2)=(x+2)(2x+2)=2(x+2)(x+1)