1.å·²ç¥ï¼x+y=1,xy=-1/2,å©ç¨å å¼å解æ±ï¼x(x+y)(x-y)-(x+y)çå¹³æ¹çå¼.
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.å·²ç¥ï¼a+b+c=11 æ±2a^2+2b^2+2c^2+4ab+4ac+4bcçå¼
2a^2+2b^2+2c^2+4ab+4ac+4bc
=2(a+b+c)^2
=2*11^2
=242
3.ï¼2000ä¸æ¬¡æ¹-2*2001äºæ¬¡æ¹-1999ï¼/2001ä¸æ¬¡æ¹+2001äºæ¬¡æ¹-2002
设2001为x.
å[(x-1)^3-2x^2-x+2]/x^3+x^2-x-1
4.1.2xçå¹³æ¹-7xy-22yçå¹³æ¹
ï¼æ³¨ï¼X^2代表Xçå¹³æ¹ï¼
2x^2-7xy-22y^2=(2x-11y)(x+2y);
5.1999xçå¹³æ¹-(1999çå¹³æ¹-1)-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)(xçå¹³æ¹-1)(x+5)-20
ï¼X+3ï¼ï¼X+1ï¼ï¼X-1ï¼ï¼X+5ï¼-20
ï¼X^2+4X+3ï¼ï¼X^2+4X-5ï¼-20
ï¼X^2+4Xï¼^2-2ï¼X^2+4Xï¼-35
ï¼X^2+4X+5ï¼ï¼X^2+4X-7)
7.å·²ç¥ä¸è§å½¢ABCçä¸éé¿åå«ä¸ºabcè¯å©ç¨å解å å¼è¯´æå¼åbçå¹³æ¹-açå¹³æ¹+2ac-cçå¹³æ¹
b^2-a^2+2ac-c^2=b^2-(a-c)^2=(b+a-c)(b-a+c)(å¹³æ¹å·®å
¬å¼)
å 为a,b,cæ¯ä¸è§å½¢ä¸è¾¹,æ ¹æ®ä¸¤è¾¹ä¹å大äºç¬¬ä¸è¾¹,æ以ä¸å¼æ¯æ£æ°.
8.æ±æ¹ç¨6xy+4x-9y-7=o çæ´æ°è§£
6xy+4x-9y-7=0
6xy-9y+4x-6-1=0
3y(2x-3)+2(2x-3)=1
(3y+2)(2x-3)=1
å½3y+2=0æ¶,æ æ´æ°è§£,å 为æ¤æ¶yä¸æ¯æ´æ°ï¼
å½3y+2â 0æ¶,2x-3=1/(3y+2)
æ¥ä¸æ¥åæ
åµè®¨è®ºï¼å½y为å°äº-1çæ´æ°æ¶ï¼å³ä»-2å¼å§ï¼,é£ä¹3y+22,ä¹å°±æ¯è¯´1/(3y+2)æ¤æ¶ä¹ä¸å¯è½ä¸ºæ´æ°,é£ä¹è§£åºçxä¹ä¸å¯è½ä¸ºæ´æ°,æ¤æ¶ä¹æ æ´æ°è§£.
å½y=-1æ¶,3y+2=-1,é£ä¹1/(3y+2)=-1,å³ 2x-3=-1,åx=1,æ以æ´æ°è§£ä¸ºï¼1,-1ï¼
温馨提示:答案为网友推荐,仅供参考