a*b=a(1-b)若a,b是方程x的平方减x加四分之一m等于0且m<0的两根,则b*b-a*a的 设a.b是方程x的平方加x减2017等于0的两个实数根,则a...
a*b=a(1-b)\u82e5a,b\u662f\u65b9\u7a0bx\u7684\u5e73\u65b9\u51cfx\u52a0\u56db\u5206\u4e4b\u4e00m\u7b49\u4e8e0\u4e14m<0\u7684\u4e24\u6839,\u5219b*b-a*a\u7684\u7531\u516c\u5f0f\uff1aa*b=a(1-b)\uff0c\u5f97\uff1aa*a=a(1-a)=a-a^2,b*b=b(1-b)=b-b^2,b*b-a*b=b-b^2-(a-a^2)=a^2-b^2-a+b=(a-b)(a+b-1),\u7531\u97e6\u8fbe\u5b9a\u7406\u5f97\uff1aa+b=1 ,\u2234 b*b-a*b=(a-b)(a+b-1),=(a-b)(1-1)=0.
\u89e3\uff1a
x=a\u4ee3\u5165\u65b9\u7a0b\uff0c\u5f97\uff1aa²+a-2017=0
a²=2017-a
\u7531\u97e6\u8fbe\u5b9a\u7406\u5f97\uff1aa+b=-1
a²+2a+b
=2017-a+2a+b
=a+b+2017
=-1+2017
=2016
a*a=a(1-a)=a-a^2,
b*b=b(1-b)=b-b^2,
b*b-a*b=b-b^2-(a-a^2)=a^2-b^2-a+b=(a-b)(a+b-1),
由韦达定理得：
a+b=1 , ab=m/4,
b*b-a*b
=(a-b)(a+b-1),
=(a-b)(1-1)
=0.
绛旓細鐢卞叕寮忥細a*b=a(1-b)锛屽緱锛歛*a=a(1-a)=a-a^2,b*b=b(1-b)=b-b^2,b*b-a*b=b-b^2-(a-a^2)=a^2-b^2-a+b=(a-b)(a+b-1),鐢遍煢杈惧畾鐞嗗緱锛歛+b=1 , ab=m/4,b*b-a*b =(a-b)(a+b-1),=(a-b)(1-1)=0.
绛旓細2 鏄鐨 (a*a)+(b*b)=a(1+b)+b(1+a)=a+b+2ab鍥犱负a+b=0锛屾墍浠=2ab 鍏朵粬鐨勪笉涓瀹1鐨勮瘽a锛1-b锛涓嶇瓑浜巄锛1-a锛3鐨勮瘽a锛1-b锛=0锛岄偅涔坅=0鎴栬卋=1锛屼笉涓瀹氭槸a=0
绛旓細鍗筹紙1*2锛*3=2 3. a*b=a(1-b)=a-ab ,b*a =b(1-a) =b-ab a*b鈮燽*a 4.a*b=a(1-b)=a-ab=a,鎵浠1鍜4姝ｇ‘銆
绛旓細绛旓細杩欓噷瑙勫畾鐨*琛ㄧず杩愮畻瑙勫緥鐨勭鍙凤紝涓嶈兘绛夊悓浜庢甯哥殑涔樺彿锛屼絾绫讳技鍔犲噺涔橀櫎绗﹀彿銆傝繍绠楄寰嬫槸锛a*b=ab+a-b 閭ｄ箞锛氬師寮=a*b+(b-a)*b =[ ab+a-b] + [ (b-a)b+(b-a)-b ]=ab+a-b+b^2-ab+b-a-b =b^2-b
绛旓細杩欎釜瀹氫箟杩愮畻鏃涓涓彇灏忓嚱鏁般傜敱浜庢寚鏁板嚱鏁板煎煙鎭掑ぇ浜庨浂銆備笖2鐨剎娆℃柟鍜2鐨-x娆℃柟浜掍负鐩稿弽鏁般傚垯鍏跺煎煙涓猴紙0锛1銆.鍗婂紑鍗婇棴銆傚綋x=0鏃躲傚彇寰1.銆傚笇鏈涢噰绾
绛旓細a+b=2ab 锛熻繖鏄笉瀹氭柟绋嬶紝涓嶅畾鏂圭▼涓鑸湁鏃犳暟缁勮В銆傚鏋滃鏈煡鏁扮殑鍊奸檺鍒惰寖鍥达紝鍏惰В鍒欏彲鑳芥湁闄愮殑銆傝В:a+b=2ab 2ab鈥b=a (2a鈥1)b =a b =a /(2a 鈥1)a=0鏃锛宐=0锛a=1鏃讹紝b=1 杩欐槸杩欎釜涓嶅畾鏂圭▼鐨勪袱缁勬暣鏁拌В銆
绛旓細鏍规嵁棰樻剰鍙煡锛4*5=4+5-1=8锛岃繖鏍凤紝(4*5)*x=8*x=8+x-1=7+x锛岃屽凡鐭ュ畠绛変簬10锛岀敱姝ゅ彲鐭ワ紝x=10-7=3銆
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