Question

1.已知1/x-1/y=3 求代数式2x-14xy-2y/x-2xy-y的值

Answer 1

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=4

2.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)