初二上2道分解因式题 y2+y+¼ [2x+y]²-[x+2y]²第一道y2+y+¼第二[2x+y]²-[x+2y]²

初二上2道分解因式题 y2+y+¼ [2x+y]²-[x+2y]²第一道y2+y+¼第二[2x+y]²-[x+2y]²
初二上2道分解因式题 y2+y+¼ [2x+y]²-[x+2y]²
第一道y2+y+¼
第二[2x+y]²-[x+2y]²

初二上2道分解因式题 y2+y+¼ [2x+y]²-[x+2y]²第一道y2+y+¼第二[2x+y]²-[x+2y]²
y²+y+¼=(y+1/2)²
(2x+y)²-(x+2y)²
=[(2x+y)+(x+2y)][(2x+y)-(x+2y)]
=(3x+3y)(x-y)
=3(x+y)(x-y)

y2+y+¼=(y+1/2)²
[2x+y]²-[x+2y]²=(2x+y+x+2y)(2x+y-x-2y)=3(x+y)(x-y)

y2+y+¼ = y^2 + 2(1/2)y + (1/2)^2 = (y+1/2)^2
[2x+y]²-[x+2y]² = (2x+y + x+2y)(2x+y-x-2y)= (3x+3y)(x-y)=3(x+y)(x-y)

第一道：y2+y+1/4=y2+1/2*y*1/4=(y+1/2)2 完全平方公式
第二道：原试=(2x+y+x+2y)(2x+y-x-2y)=3(x+y)(x-y)=3(x2-y2) 平方差公式