Question

当a=5.5,b=4.5时[-1/4(a+b)^2(a-b)]^3*[-4（a+b）*（a-b）^2]^3

先化解在代入ab的值

Answer 1

∵π/2<b<a<3π/4,
∴0<a-b<π/4,π<a+b<3π/2.
∵cos(a-b)=12/13,sin(a+b)=-3/5,
∴sin(a-b)=√(1-cos2(a-b))=5/13,
cos(a+b)=-√(1-sin2(a+b))=-4/5.
故sin2a=sin[(a+b)+(a-b)]
=sin(a+b)cos(a-b)+cos(a+b)sin(a-b)
=(-3/5)(12/13)+(-4/5)(5/13)
=-56/65,
sin2b=sin[(a+b)-(a-b)]
=sin(a+b)cos(a-b)-cos(a+b)sin(a-b)
=(-3/5)(12/13)-(-4/5)(5/13)
=-16/65.
相信我，没错的

Answer 2

=（-1/4*100）^3*(-40)^3
=(-25)^3*(-40)^3
=(25*40)^3
=(1000)^3
=((10)^3)^3
=10^9

Answer 3

[-1/4(a+b)^2(a-b)]^3*[-4（a+b）*（a-b）^2]^3
=[-1/(4*10^2*1)]^3*[-4*10*1^2]^3
= -1/400^3 * (-40)^3
=1/1000

Answer 4

a+b=10,a-b=1
代入得[-1/4x100x1]^3[-4x10x1]^3=(-25)^3(-40)^3=(25x40)^3=10^9

Answer 5

原式=[-1/4*10^2*1]^3*[-4*10*1^2]^3
=(-25)^3*(-40)^3
=[(-25)*(-40)]^3
=1*10^9