Question

化简：⑴（a/3+2)²-(a/3-2)² ⑵(a+2b+c)² ⑶(a-b+c+d)(a+b+c-d)

Answer 1

⑴（a/3+2)²-(a/3-2)²

=[（a/3+2)+(a/3-2)]*[（a/3+2)-(a/3-2)]

=2/3a*4=8/3a

⑶(a-b+c+d)(a+b+c-d)

= [(a+c)-(b-d)][(a+c)+(b-d) ]

=(a+c)²-(b-d)²