当x分之1—y分之1=3时,求[x-2xy-y}分之{2x+3xy-2y
问题描述：
当x分之1—y分之1=3时,求[x-2xy-y}分之{2x+3xy-2y
答
1、
f(x)=4[1-cos2(π/4+x)]/2-2√3cos2x-1
=2[1-cos(π/2+2x)]-2√3cos2x-1
=2+2sin2x-2√3cos2x-1
=4(sin2x*1/2-cos2x*√3/2)+1
=4(sin2xcosπ/3-cos2xsinπ/3)+1
=4sin(2x-π/3)-1
π/4π/6所以1/21所以值域[1,3]
2、
sinx增区间是(2kπ-π/2,2kπ+π/2)
π/6所以这里是(π/6,π/2)
π/6π/2π/4
所以
增区间(π/4,5π/12)
减区间(5π/12,π/2)
答
1/x-1/y=(y-x)/xy=3
y-x=3xy
x-y=-3xy
原式=[2(x-y)+3xy]/[(x-y)-2xy]
=[2(-3xy)+3xy]/[(-3xy)-2xy]
=-3xy/(-5xy)
=3/5