帮我解2道数学题

第一题:原式=[(ab+1)-(ab-1)][(ab+1)+(ab-1)]
=2(2ab)
=4ab
第2题比较复杂,不打了

（ab+1)^2-(ab-1)^2
=[（ab+1）+（ab-1）]*[（ab+1）-（ab-1）]
=2ab*2=4ab

(2k-y)^2-4(k-y)(k+2y)
=[(2k-y)-2(k-y)]^2-[2(k+2y)]^2
=Y^2-[2(k+2y)]^2
=[y+2(k+2y)]*[y-2(k+2y)]
=-(2k+5y)*(2k+3y)

（ab+1)^2-(ab-1)^2
=[(ab+1)+(ab-1)]×[(ab+1)-(ab-1)]
=(ab+1+ab-1)×(ab+1-ab+1)
=2ab×2
=4ab

(2k-y)^2-4(k-y)(k+2y)
=(4k^2+y^2-4ky)-4(k^2+2ky-ky-2y^2)
=4k^2+y^2-4ky-4(k^2+ky-2y^2)
=4k^2+y^2-4ky-4k^2-4ky+8y^2
=9y^2-8ky

[(ab+1)+(ab-1)]*[(ab+1)-(ab-1)]=2ab*2=4ab

4k^2-4ky+y^2-4k^2-4ky+8y^2=-8ky+9y^2

(ab+1)^2-(ab-1)^2
=(ab+1+ab-1)(ab+1-ab+1)
=2ab*2
=4ab

(2k-y)^2-4(k-y)(k+2y)
=(4k''2-4ky+y''2)-4(k''2+ky-2y''2)
=4k''2-4ky+y''2-4k''2-4ky+8y''2
=-8xy+9y''2