Question

2(a+b)(a-b)-(a+b)(a+b)+(a-b)(a-b) 其中a=2 b=2/1

Answer 1

原式=2(a^2-b^2)-(a^2+2ab+b^2)+(a^2-2ab+b^2)=2(a^2-b^2)-4ab
b=2/1?
若是，则结果为-16

Answer 2

2(a+b)(a-b)-(a+b)(a+b)+(a-b)(a-b)
=2a²-2b²-a²-2ab-b²＋a²-2ab＋b²
=2a²-2b²-4ab
=2×2²-2×(2分之1)²-4×2×2分之1
=8-2分之1-4
=4-2分之1
=2分之7