Question

（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)其中a=-4分之1 b=0.4

Answer 1

（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)
=4a^2-2ab-b^2+a^2-b^2+2ab+3a^2-ab+b^2
=8a^2-ab-b^2
将a=-1/4，b=0.4=2/5代入上式，得8*(-1/4)^2-(-1/4)*2/5-(2/5)^2=8*1/16+1/10-4/25=1/2+1/10-4/25=11/25

Answer 2

当a=-1/4 ，b=0.4
（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)
=4a²-2ab-b²+a²-b²+2ab+3a²-ab+b²
=8a²-ab-b²
=8×（-1/4）²-（-1/4）×0.4-0.4²
=0.5+0.1-0.16=0.44

Answer 3

（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)
=4a²-2ab-b²+a²-b²+2ab+3a²-ab+b²
=8a²-ab-b²
=8x(-1/4)²+1/4x0.4-(0.4)²
=1/2+0.1-0.16
=0.44

Answer 4

（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)
=4a^2-2ab-b^2+a^2-b^2+2ab+3a^2-ab+b^2
=4a^2+3a^2+a^2-2ab+2ab-ab+b^2-b^2-b^2
=8a^2-ab-b^2
=8*(1/4)^2-(-1/4)*0.4-0.4^2
=8*1/16-(-1/4*2/5)-(2/5)^2
=1/2+1/10-4/25
=25/50+5/50-8/50
=22/50
=11/25

Answer 5

解：（4a^2-2ab-b^2）-（-a^2+b^2-2ab）+（3a^2-ab+b^2)=8a^2-ab-b^2
其中a=-4分之1 b=0.4
原式=8x1/16-1/4x0.4-0.4^2=1/2-0.1-0.4x0.4=0.4x(1-0.4)=0.24