Question

初一完全平方计算题

1.（2a+1)�0�5(2a-1)�0�52.(x-2y)�0�5+2(x+2y)(x-2y)+(x+2y)�0�53.(a+b+2c)(a+b-2c)4.(x+2y-z)(x-2y+z)5.(a+b+c)�0�56.（x-2y+1/3）�0�5

Answer 1

1.（2a+1)�0�5(2a-1)�0�5=[(2a+1)(2a-1)]�0�5=(4a�0�5-1)�0�5=14a4-8a�0�5+12.(x-2y)�0�5+2(x+2y)(x-2y)+(x+2y)�0�5=[(x-2y)+(x+2y)]�0�5=4x�0�53.(a+b+2c)(a+b-2c)=(a+b)�0�5-4c�0�5=a�0�5+2ab+b�0�5-4c4.(x+2y-z)(x-2y+z)=x�0�5-(2y-z)�0�5=x�0�5-4y�0�5+4yz-z�0�55.(a+b+c)�0�5=(a+b)�0�5+2(a+b)c+c�0�5=a�0�5+b�0�5+c�0�5+2ab+2ac+2bc6.（x-2y+1/3）�0�5=(x-2y)+2/3(x-2y)+1/9=x�0�5-4xy+4y�0�5+2/3x-4/3y+1/3

Answer 2

你是要分解因式吗？