Question

1.(x²+y²）（x²-y²）-（x+y）²（x-y）²，其中x=4，y=1/4

2.（2a-1/2b）（1/2b+2a)(1/4b²+4a²），其中a=-1，b=-2

Answer 1

1.
原式= (x²+y²)(x²-y²) - (x²-y²)² = (x²-y²)(x²+y² - x² + y²) = 2y²(x²-y²) = 2(1/4)²[4² - (1/4)²]
= 255/128

2. 原题不清楚，假定为(2a - b/2)(b/2 + 2a)(b²/4 + 4a²)
= (4a² -b²/4)(4a² + b²/4)
= 16a^4 -b^4/16
= 16(-1)^4 - (-2)^4/16
= 16 - 1
= 15

Answer 2

1.(x²+y²）（x²-y²）-（x+y）²（x-y）²
=x^4-y^4-[(x+y)(x-y)^2
=x^4-y^4-x^2+2x^4y^2-y^4
=2x^2y^2
=2*(4*1/4)^2
=2
2.（2a-1/2b）（1/2b+2a)(1/4b²+4a²）
=(4a^2-1/4b^2)(1/4b²+4a²）
=16a^4-1/16b^4
=16*1-1/16*16
=15