初二数学化简求值题 初二数学题30道因式分解30道化简求值(有
\u6c4250\u9053\u521d\u4e8c\u4e0b\u518c\u6570\u5b66\u56e0\u5f0f\u5206\u89e3\u5316\u7b80\u6c42\u503c\u98981.\u5df2\u77e5\uff1ax+y=1,xy=-1/2,\u5229\u7528\u56e0\u5f0f\u5206\u89e3\u6c42\uff1ax(x+y)(x-y)-(x+y)\u7684\u5e73\u65b9\u7684\u503c.
x(x+y)(x-y)-x(x+y)^2
=x(x+y)(x-y-x-y)
=x(x+y)(-2y)
=-2xy(x+y)
=-2*[-1/2]*1
=1
2.\u5df2\u77e5\uff1aa+b+c=11 \u6c422a^2+2b^2+2c^2+4ab+4ac+4bc\u7684\u503c
2a^2+2b^2+2c^2+4ab+4ac+4bc
=2(a+b+c)^2
=2*11^2
=242
3.\uff082000\u4e09\u6b21\u65b9-2*2001\u4e8c\u6b21\u65b9-1999\uff09/2001\u4e09\u6b21\u65b9+2001\u4e8c\u6b21\u65b9-2002
\u8bbe2001\u4e3ax.
\u5219[(x-1)^3-2x^2-x+2]/x^3+x^2-x-1
4.1.2x\u7684\u5e73\u65b9-7xy-22y\u7684\u5e73\u65b9
\uff08\u6ce8\uff1aX^2\u4ee3\u8868X\u7684\u5e73\u65b9\uff09
2x^2-7xy-22y^2=(2x-11y)(x+2y);
5.1999x\u7684\u5e73\u65b9-(1999\u7684\u5e73\u65b9-1)-1999
1999x^2-(1999^2-1)x-1999
=1999x^2-1999^2x+x-1999
=1999x(x-1999)+(x-1999)
=(x-1999)(1999x+1)
6.(x+3)(x\u7684\u5e73\u65b9-1)(x+5)-20
\uff08X+3\uff09\uff08X+1\uff09\uff08X-1\uff09\uff08X+5\uff09-20
\uff08X^2+4X+3\uff09\uff08X^2+4X-5\uff09-20
\uff08X^2+4X\uff09^2-2\uff08X^2+4X\uff09-35
\uff08X^2+4X+5\uff09\uff08X^2+4X-7)
7.\u5df2\u77e5\u4e09\u89d2\u5f62ABC\u7684\u4e09\u904d\u957f\u5206\u522b\u4e3aabc\u8bd5\u5229\u7528\u5206\u89e3\u56e0\u5f0f\u8bf4\u660e\u5f0f\u5b50b\u7684\u5e73\u65b9-a\u7684\u5e73\u65b9+2ac-c\u7684\u5e73\u65b9
b^2-a^2+2ac-c^2=b^2-(a-c)^2=(b+a-c)(b-a+c)(\u5e73\u65b9\u5dee\u516c\u5f0f)
\u56e0\u4e3aa,b,c\u662f\u4e09\u89d2\u5f62\u4e09\u8fb9,\u6839\u636e\u4e24\u8fb9\u4e4b\u548c\u5927\u4e8e\u7b2c\u4e09\u8fb9,\u6240\u4ee5\u4e0a\u5f0f\u662f\u6b63\u6570.
8.\u6c42\u65b9\u7a0b6xy+4x-9y-7=o \u7684\u6574\u6570\u89e3
6xy+4x-9y-7=0
6xy-9y+4x-6-1=0
3y(2x-3)+2(2x-3)=1
(3y+2)(2x-3)=1
\u5f533y+2=0\u65f6,\u65e0\u6574\u6570\u89e3,\u56e0\u4e3a\u6b64\u65f6y\u4e0d\u662f\u6574\u6570\uff1b
\u5f533y+2\u22600\u65f6,2x-3=1/(3y+2)
\u63a5\u4e0b\u6765\u5206\u60c5\u51b5\u8ba8\u8bba\uff1a\u5f53y\u4e3a\u5c0f\u4e8e-1\u7684\u6574\u6570\u65f6\uff08\u5373\u4ece-2\u5f00\u59cb\uff09,\u90a3\u4e483y+22,\u4e5f\u5c31\u662f\u8bf41/(3y+2)\u6b64\u65f6\u4e5f\u4e0d\u53ef\u80fd\u4e3a\u6574\u6570,\u90a3\u4e48\u89e3\u51fa\u7684x\u4e5f\u4e0d\u53ef\u80fd\u4e3a\u6574\u6570,\u6b64\u65f6\u4e5f\u65e0\u6574\u6570\u89e3.
\u5f53y=-1\u65f6,3y+2=-1,\u90a3\u4e481/(3y+2)=-1,\u5373 2x-3=-1,\u5219x=1,\u6240\u4ee5\u6574\u6570\u89e3\u4e3a\uff081,-1\uff09
1.a^4-4a+3
2.(a+x)^m+1*(b+x)^n-1-(a+x)^m*(b+x)^n
3.x^2+(a+1/a)xy+y^2
4.9a^2-4b^2+4bc-c^2
5.(c-a)^2-4(b-c)(a-b)
\u7b54\u68481.\u539f\u5f0f=a^4-a-3a+3=(a-1)(a^3+a^2+a-3)
2.[1-(a+x)^m][(b+x)^n-1]
3.(ax+y)(1/ax+y)
4.9a^2-4b^2+4bc-c^2=(3a)^2-(4b^2-4bc+c^2)=(3a)^2-(2b-c)^2=(3a+2b-c)(3a-2b+c)
5.(c-a)^2-4(b-c)(a-b)
= (c-a)(c-a)-4(ab-b^2-ac+bc)
=c^2-2ac+a^2-4ab+4b^2+4ac-4bc
=c^2+a^2+4b^2-4ab+2ac-4bc
=(a-2b)^2+c^2-(2c)(a-2b)
=(a-2b-c)^2
绛旓細鏈熶腑鑰冭瘯涓撻澶嶄範锛3锛夊垎寮忕殑鍖栫畝姹傚 1銆侊紙2011鏈熶腑锛夊厛鍖栫畝锛屽啀姹傚 梅 锛屽叾涓瓁= 2銆侊紙2010鏈熶腑锛夎绠 梅 3銆侊紙06姝︽眽锛夊厛鍖栫畝锛屽啀姹傚硷細4.(07姝︽眽)鍖栫畝姹傚硷細 锛屽叾涓瓁锛2銆5.锛08姝︽眽锛夊厛鍖栫畝锛屽啀姹傚硷細 锛屽叾涓 锛6.锛09姝︽眽锛夊厛鍖栫畝锛屽啀姹傚硷細 锛屽叾涓 锛7.锛10姝︽眽锛夊厛鍖栫畝锛屽啀姹傚...
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绛旓細a/(a+b)-a²/(a²-b²)=a/(a+b)-a²/(a+b)(a-b)=a(a-b)/(a+b)(a-b)-a²/(a+b)(a-b)=(a²-ab-a²)/(a+b)(a-b)=-ab/(a+b)(a-b)
绛旓細1.宸茬煡锛歺+y=1,xy=-1/2,鍒╃敤鍥犲紡鍒嗚В姹傦細x(x+y)(x-y)-(x+y)鐨勫钩鏂圭殑鍊.x(x+y)(x-y)-x(x+y)^2 =x(x+y)(x-y-x-y)=x(x+y)(-2y)=-2xy(x+y)=-2*[-1/2]*1 =1 2.宸茬煡锛歛+b+c=11 姹2a^2+2b^2+2c^2+4ab+4ac+4bc鐨勫 2a^2+2b^2+2c^2+4ab+...
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绛旓細瑙o細鏂圭▼鍙樺舰涓猴細x-鏍瑰彿涓嬶紙xy锛=鏍瑰彿涓嬶紙xy锛+3y x-2鍊嶆牴鍙蜂笅锛坸y锛-3y=0 锛堟牴鍙穢锛塣2-2鍊嶆牴鍙穢*鏍瑰彿y-3鍊嶏紙鏍瑰彿y锛塣2=0 (鏍瑰彿x-3鍊嶆牴鍙穣锛夛紙鏍瑰彿x+鏍瑰彿y锛=0 鍥犱负鏍瑰彿x锛0锛屾牴鍙穣锛0 鎵浠ユ牴鍙穢+鏍瑰彿y锛0 鍒欐牴鍙穢-3鍊嶆牴鍙穣=0 鏍瑰彿x=3鍊嶆牴鍙穣 鍗硏=9y 鎺ヤ笅鏉ュ皢x=9y浠e叆...