如图,点B在射线AE上,∠CAE=∠DAE,∠CBE=∠DBE.求证:AC=AD.
(1)分解因式:x2+2x+1= .(2)若∠α=40°,则∠α的余角是 .
分解因式:(1)(m+2n)2﹣(m﹣n)2(2)4(a+b)﹣(a+b)2﹣4
分解因式:a2﹣2ab+b2﹣c2.
(1)﹣8a2b+2a3+8ab2; (2)(x+y)2+2(x+y)+1;(3)x2(x﹣y)+(y﹣x); (4)x2﹣2xy+y2﹣9.