求解一初中数学因式分解题 初中数学因式分解题
\u521d\u4e2d\u6570\u5b66\u56e0\u5f0f\u5206\u89e3\u4e00\u9898\u6c42\u89e3\u56e0\u4e3a:X3-Y3-Z3=3XYZ
\u6240\u4ee5:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0
(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0
\u6240\u4ee5:1. X-Y-Z=0
2. X2+Y2+Z2+XY+XZ-YZ=0
\u5f531\u65f6:X-Y-Z=0
X2=(Y+Z)2=2(Y-Z)
\u6240\u4ee5:Y=X(X+2)/4 Z=X(2-X)/4
\u6240\u4ee5:XY+YZ+ZX=X2(20-X2)/16
\u5f532\u65f6:X2+Y2+Z2+XY+XZ-YZ=0
X2=2(Y-Z)
XY+YZ+ZX=2YZ-X2-Y2-Z2
=-(Y-Z)2-x2
=(Z-Y)(Y-Z+2)
\u7b2c\u4e00\u9898\u5229\u7528\u5e73\u65b9\u5dee\u516c\u5f0f\uff0c\u9996\u5148\u628a9\u548c16\u8f6c\u5316\u4e3a3²\u548c4²\u7136\u540e\u4ee3\u5165\u5f0f\u5b50\uff0c\u5c31\u4f1a\u53d8\u6210\uff086x-3y\uff09²-\uff084x+8y\uff09²\uff0c\u7136\u540e\u4f7f\u7528\u5e73\u65b9\u5dee\u516c\u5f0f\u5c31\u884c\u4e86\u3002\uff08a+b\uff09\uff08a-b\uff09=a²-b²\u3002
\u6700\u540e\u5f97\u5230\uff0810x+5y\uff09\uff082x-11y\uff09=5\uff082x+y\uff09\uff082x-11y\uff09\u3002
\u7b2c\u4e8c\u9898\uff0c\u6ce8\u610f\u5230x\uff082-x\uff09\u5c31\u662f-\uff08x²-2x\uff09\uff0c\u5f0f\u5b50\u8f6c\u5316\u4e3a\uff08x²-2x\uff09²+2\uff08x²-2x\uff09+1\uff0c\u6beb\u65e0\u7591\u95ee\u662f121\u5f62\u5f0f\u7684\u5e73\u65b9\uff0c\u6240\u4ee5\u662f\uff08x²-2x+1\uff09²\u3002\u7136\u540e\u53d1\u73b0\uff08x²-2x+1\uff09\u53ef\u4ee5\u5316\u7b80\u4e3a\uff08x-1\uff09²\u3002
\u6700\u540e\u5f97\u5230\uff08x-1\uff09^4.
\u7b2c\u4e09\u9898\uff0c\u9996\u5148\u63d0\u53d6x^n*y\uff0c\u5f0f\u5b50\u8f6c\u5316\u4e3ax^n*y\uff08x²-y²\uff09\uff0c\u7136\u540e\u4f7f\u7528\u5e73\u65b9\u5dee\u516c\u5f0f\u3002
\u6700\u540e\u5f97\u5230x^n*y\uff08x+y\uff09\uff08x-y\uff09\u3002
\u4ee5\u4e0a*\u8868\u793a\u4e58\u6cd5\uff0c^\u8868\u793a\u6b21\u65b9\uff0c\u6bd4\u59822^5\u5c31\u662f\u4e8c\u7684\u4e94\u6b21\u65b9\u3002
1-1/2方=（1+1/2）（1-1/2）
因此，原式=（1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)...(1-1/10)(1+1/10)
=1/2*3/2*2/3*4/3*3/4*5/4*...*9/10*11/10（倒数乘积是1，就首尾相乘就得）
=1/2*11/10=11/20
解答完毕
（1-1/2²）（1-1/3²）（1-1/4²）……（1-1/10²）
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)...(1-1/10)(1+1/10)
=1/2*3/2*2/3*4/3*3/4*5/4*...*9/10*11/10
=1/2*11/10=11/20
1 - 1/2² =1² -（1/2）² =(1+1/2)(1-1/2)=3/2 *1/2=3/4
同理原题 =3/4 * 8/9 *15/16 * 24/25 * 35/36 *48/49 * 63/64 * 80/81 * 99/100
=11/20
（1-1/2²）=3/4 (1-1/3²)=8/9..........................（1-1/10²）=99/100
原式=3/4×8/9×........×99/100
=11/20
绛旓細1.X鐨勫钩鏂-4 =(x+2)(x-2)2.9-Y鐨勫钩鏂=(3+y)(3-y)3.1-a鐨勫钩鏂=(1+a)(1-a)4.4X鐨勫钩鏂-Y鐨勫钩鏂=(2x+y)(2x-y)5.a鐨勫钩鏂-9鍒嗕箣涓X鐨勫钩鏂=(a+1/3x)(a-1/3x)6.36-m鐨勫钩鏂=(6+m)(6-m)7.4X鐨勫钩鏂-9Y鐨勫钩鏂=(3x+2y)(3x-2y)8.0.81a鐨勫钩鏂-16b=(0.9a+4b...
绛旓細4.27y²+18y+3 =3(9y²+6y+1)=3(3y+1)²
绛旓細绗竴棰鍒╃敤骞虫柟宸叕寮忥紝棣栧厛鎶9鍜16杞寲涓3²鍜4²鐒跺悗浠ｅ叆寮忓瓙锛屽氨浼氬彉鎴愶紙6x-3y锛²-锛4x+8y锛²锛岀劧鍚庝娇鐢ㄥ钩鏂瑰樊鍏紡灏辫浜嗐傦紙a+b锛夛紙a-b锛=a²-b²銆傛渶鍚庡緱鍒帮紙10x+5y锛夛紙2x-11y锛=5锛2x+y锛夛紙2x-11y锛夈傜浜岄锛屾敞鎰忓埌x锛2-x锛夊氨鏄-锛坸...
绛旓細=锛峹2y(x2锛峹y锛媦2)(3)xn锛3xn锛1锛媥n锛2=xn锛2路x2锛媥n锛2路3x锛媥n锛2路1 =xn锛2(x2锛3x锛1)(4)5(x锛峺)3锛10(y锛峹)2=5(x锛峺)3锛10(x锛峺)2 =5(x锛峺)2(x锛峺锛2)(5)m(5ax锛媋y锛1)锛峬(3ax锛峚y锛1)=m[(5ax锛媋y锛1)锛(3ax锛峚y锛1)]=m路(5ax锛媋y锛...
绛旓細a^2-b^2-2a+1 =[a-1]^2-b^2 =[a-b-1][a+b-1](ax+by)^2+(bx-ay)^2 =a^2x^2+b^2y^2+2abxy+b^2x^2+a^2y^2-2abxy =x^2[a^2+b^2]+y^2[b^2+a^2]=[a^2+b^2][x^2+y^2](a+b)^2-4(a^2-b^2)+4(a-b)^2 =[a+b]^2-4[a+b][a-b]+4...
绛旓細2ax^2锛3x锛2ax锛3锛(x+1)(2ax-3)x(y锛2)锛峹锛峺锛1锛(x-1)(y+1)(x^2锛3x)锛(x锛3)^2锛(x-3)(2x-3)9x^2锛66x锛121锛(3x-11)^2 8锛2x^2锛2(2-x)(2+x)x^4锛1锛(x-1)(x+1)(x^2+1)x^2锛4x锛峹y锛2y锛4锛(x+2)(x-y+2)4x^2锛12x锛5锛(2x-1)...
绛旓細瑙ｏ細m^2=n+2 n=m^2-2鈥︹︹︼紙1锛塶^2=m+2鈥︹︹︼紙2锛夋妸锛1锛変唬鍏ワ紙2锛夛細(m^2-2)^2=m+2 m^2(m+2)(m-2)-(m-2)=0 m-2=0 m=2 鎴 m^3+2m^2-1=0 m^3+m^2+m^2-1=0 m^2(m+1)+(m+1)(m-1)=0 (m+1)(m^2+m-1)=0 m+1=0 m=-1 鎴栵細m^2+m...
绛旓細2銆佸崄瀛楃浉涔樻硶鐨勭敤澶:(1)鐢ㄥ崄瀛楃浉涔樻硶鏉鍒嗚В鍥犲紡銆(2)鐢ㄥ崄瀛楃浉涔樻硶鏉瑙ｄ竴鍏冧簩娆℃柟绋嬨 3銆佸崄瀛楃浉涔樻硶鐨勪紭鐐:鐢ㄥ崄瀛楃浉涔樻硶鏉瑙ｉ鐨勯熷害姣旇緝蹇,鑳藉鑺傜害鏃堕棿,鑰屼笖杩愮敤绠楅噺涓嶅ぇ,涓嶅鏄撳嚭閿欍 4銆佸崄瀛楃浉涔樻硶鐨勭己闄:1銆佹湁浜涢鐩敤鍗佸瓧鐩镐箻娉曟潵瑙ｆ瘮杈冪畝鍗,浣嗗苟涓嶆槸姣忎竴閬撻鐢ㄥ崄瀛楃浉涔樻硶鏉ヨВ閮界畝鍗曘2銆佸崄瀛楃浉涔...
绛旓細=(m+1)(m2-m+1)(m6+1+m3)(m6+1-m3)渚2鍒嗚В鍥犲紡锛歺4+5x3+15x-9 瑙ｆ瀽鍙牴鎹郴鏁扮壒寰佽繘琛屽垎缁 瑙ｅ師寮=锛坸4-9锛+5x3+15x =(x2+3)(x2-3)+5x(x2+3)=(x2+3)(x2+5x-3)闄勶細浠呬緵鍙傝 绗4璇 鍥犲紡鍒嗚В 銆栫煡璇嗙偣銆楀洜寮忓垎瑙ｅ畾涔夛紝鎻愬彇鍏洜寮忋佸簲鐢ㄥ叕寮忔硶銆佸垎缁勫垎瑙ｆ硶銆佷簩娆′笁椤瑰紡...
绛旓細1.宸茬煡澶氶」寮2x²-y²-xy-x-2y-1鐨勪竴涓鍥犲紡鏄(x-y-1),鍙︿竴涓洜寮忎负Ax+By+C.姹(A+B)路C鐨勫.瑙ｏ細鐢遍鎰忓緱锛(x-y-1)脳(Ax+By+C)=Ax²-Axy-Ax+Bxy-By²-By+Cx-Cy-C =Ax²-By²+(B-A)xy+(C-A)x+(-B-C)y-C =2x²-y²...
