初一的因式分解，急
\u521d\u4e00\u56e0\u5f0f\u5206\u89e31.\uff08x^2-3)^2+2(x^2-3)(x-3)+(x-3)^2
=(x^2-3+x-3)^2
=(x^2+x-6)^2
=(x+3)^2(x-2)^2
2.(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x)+22(x^2+8x)+105+15
=(x^2+8x)+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
3.x^2+y^2+4x+2y+3\u2014\u2014\u9898\u76ee\u662f\u4e0d\u662f\u5199\u9519\u4e86\u5440
x^2-y^2+4x+2y+3
=(x^2+4x+4)-(y^2-2y+1)
=(x+2)^2-(y-1)^2
=(x+2+y-1)(x+2-y+1)
=(x+y+1)(x-y+3)
4.(a^2+a+1)^2+2(a^2+a+1)(a^2+a-1)+(a^2+a-1)^2
=(a^2+a+1+a^2+a-1)^2
=(2a^2+2a)^2
=[2a(a+1)]^2
=4a^2(a+1)^2
5.(x^2+4)^2-16x^2
=(x^2+4+4x)(x^2+4-4x)
=(x+2)^2(x-2)^2
6.(a+b+c)^2-(a-b-c)^2
=(a+b+c+a-b-c)(a+b+c-a+b+c)
=2a(2b+2c)
=4a(b+c)
7.25(x-y)^2-16(x+y)^2
=[5(x-y)+4(x+y)][5(x-y)-4(x+y)]
=(9x-y)(x-9y)
8.x^4-81x^2y^2
=x^2(x^2-9y^2)
=x^2(x+3y)(x-3y)
9.x^2(x-y)+y^2(y-x)
=(x^2-y^2)(x-y)
=(x+y)(x-y)^2
10.(a-b)(x-y)+(b-a)(x+y)
=(b-a)(-x+y+x+y)
=2y(b-a)
(\u4e58\u6cd5)
11.2(2m-1)^2-3(m+1)(m-1)-2(m-1)^2
=8m^2-8m+2-3m^2+3-2m^2+4m-2
=3m^2-4m+3
12.(2x+y+3t)(2x-y-3t)
=(2x)^2-(y+3t)^2
=4x^2-y^2-6ty-9t^2
13.(3a+1)^2(3a-1)^2
=[(3a+1)(3a-1)]^2
=(9a^2-1)^2
=81a^4-18a^2+1
14.(2x-y)(2x+y)-(3x-y)(4x-y)
=4x^2-y^2-(12x^2-7xy+y^2)
=-8x^2+7xy-2y^2
15.(-4x+2y)(-2y-4x)-(2x+3y)(8x-9y)
=(-4x)^2-(2y)^2-(16x^2+6xy-27y^2)
=16x^2-4y^2-16x^2-6xy+27y^2
=23y^2-6xy
16.(2a+b-c)^2
=(2a)^2+2\u00b72a(b-c)+(b-c)^2
=4a^2+4ab-4ac+b^2-2bc+c^2
=4a^2+b^2+c^2+4ab-4ac-2bc
\u89e3\uff1a
4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2 =4(2x^2-x+1)(x^2-2x+3)-[(2x^2-x+1)+(x^2-2x+3)]^2 =4(2x^2-x+1)(x^2-2x+3)-(2x^2-x+1)^2-(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3) =-(2x^2-x+1)^2-(x^2-2x+3)^2+2(2x^2-x+1)(x^2-2x+3) =-[(2x^2-x+1)^2+(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3) =-[(2x^2-x+1)-(x^2-2x+3)]^2 =-(x^2+x-2)^2 =-(x+2)^2(x-1)^2
\u795d\u5b66\u4e60\u8fdb\u6b65
2、(202+98)^2=300^2=90000
3、（184.5+15.5）^2=200^2=40000
绛旓細21,x^-15x+26=(x-2)(x-13)22,x^-25x+26=娌℃硶鍒嗚В銆23,x^-25x-26=(x-26)(x+1)24,x^+5x-6=(x+6)(x-1)
绛旓細锛1锛 锛坅+b锛²+6锛坅+b锛+9=[(a+b)+3]²=(a+b+3)²锛2锛 16 -24锛坸-y锛+9锛坸-y锛²=[4-3(x-y)]²=(4-3x+3y)²
绛旓細1銆佸畬鍏ㄥ钩鏂瑰紡锛屽舰濡傦細a^2卤2ab+b^2=(a卤b)^2 2銆佸钩鏂瑰樊鍏紡锛屽舰濡傦細a^2-b^2=(a+b)(a-b)3銆佺珛鏂瑰拰鍏紡锛屽舰濡傦細a^3+b^3=(a+b)(a^2-ab+b^2锛4銆佺珛鏂瑰樊鍏紡锛屽舰濡傦細a^3-b^3=(a-b)(a^2+ab+b^2锛5銆.鍗佸瓧鐩镐箻娉曪紝渚嬪锛歺^2-3x+2=(x-1)(x-2)6銆.鎻愬彇鍏...
绛旓細1,x^-x-20 =(x-5)(x+4)2,x^-9x+20 =(x-5)(x-4)3,x^-9x-20 =(x-3+鏍瑰彿14锛夛紙x-3-鏍瑰彿14)4,x^+9x+20 =(x+5)(x+4)5,x^-18x+32 =(x-2)(x-16)6,x^+14x-31 7,x^+18x+32 8,x^+14x-32 9,x^+19x+48 10,x^+47x-48 11,x^+13x-48 12,x^+2x...
绛旓細锛1锛6*x鐨勫钩鏂+5xy-6*y鐨勫钩鏂-6xz+4yz =(3x-2y)*(2x+3y)-2z*(3x-2y)=(3x-2y)*(2x+3y-2z)锛2锛墄鐨勫钩鏂-xy+2x+y-3 =(x^2+3x-4)-(xy+x)+(y+1)=(x-1)(x+4)-x(y+1)+(y+1)=(x-1)(x+4)-(y+1)(x-1)=(x-1)(x-y+3)...
绛旓細1.鍒嗚В鍥犲紡锛2.鍒嗚В鍥犲紡锛9x鐨勫钩鏂-6x-y鐨勫钩鏂+4y-3_=(3x-1)^2-(y-2)^2 =(3x+y-3)(3x-y+1锛塤__3.鍒嗚В鍥犲紡锛歺骞虫柟+5xy+x+3y+6y鐨勫钩鏂 =(x+2y)(x+3y)+(x+3y)=(x+3y)(x+2y+1)___5.鍒嗚В鍥犲紡锛歺鐨勫钩鏂-2X-2y鐨勫钩鏂+4y-xy =(x+y)(x-2y)-2(x+y)=(x+y)(...
绛旓細1.鍒嗚В鍥犲紡(1+y)-2x(1+y)+x(1-y) 瑙:鍘熷紡=(1+y)+2(1+y)+x(1-y)+x(1-y)-2(1+y)x(1-y)-2x(1+y) =[(1+y)+x(1-y)]-2(1+y)x(1-y)-2x(1+y) =[(1+y)+x(1-y)]-(2x) =[(1+y)+x(1-y)+2x]路[(1+y)+x(1-y)-2x] =(x-xy+2x+y+1)(x-xy-2x+y...
绛旓細1銆0.25X锛25.3+78.6-3.9锛=0.25X100=25 2銆(202+98)^2=300^2=90000 3銆侊紙184.5+15.5锛塣2=200^2=40000
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