Question

因式分解 已知a+b=2，求(a²-b²)²-8(a²+b²)

Answer 1

原式=(a+b)²(a-b)²-8(a²+b²)=4（a-b)²-8(a²+b²)=4(a²+b²-2ab)-8(a²+b²)
=-4(a²+b²+2ab)=-4(a+b)²=-4*4=-16

Answer 2

(a²-b²)²-8(a²+b²)=(a-b)²(a+b)²-8(a²+b²)=4(a-b)²-8(a²+b²)=4(a²-2ab+b²-2a²-2b²)=-4(a+b)²=-16

Answer 3

(a²-b²)²-8(a²+b²)
=(a+b)²(a-b)²-8a²-8b²
=4a²+4b²-8ab-8a²-8b²
=-4(a+b)²
=-16

Answer 4

=(a-b)^2*(a+b)^2-8[(a+b)^2-2ab]
=4[(a+b)^2-4ab]-8(4-2ab)
=4(4-4ab)-8(4-2ab)
=16-16ab-32+16ab
=-16

Answer 5

(a²-b²)²-8(a²+b²)
=[(a+b)(a-b)]²-8(a²+b²)
=[2(a-b)]²-8(a²+b²)
=4(a²-2ab+b²)-8(a²+b²)
=4a²-8ab+4b²-8a²-8b²
=-4a²-8ab-4b²
=-4(a²+2ab+b²)
=-4(a+b)²
=-4×2²
=-4×4
=-16