3、42)=123223433454=1254=58;
(1-122)(1-132)(1-142)(1-152)=1232234334544565=1265=35;
……
(1-122)(1-132)(1-142)…(1-1n2)= (用含n的代數(shù)式表示,n是正整數(shù),且n≥2).
10.(1)化簡:4a+4b5ab15a2ba2-b2.
(2)[xx鹽城] 先化簡,再求值:(1- 1x+1)xx2-1,其中x=2+1.
11.[xx曲靖] 先化簡,再求值(1a-b-ba2-b2)a2-aba2-2ab+b2,其中a,b滿足a+b-12=0.
12.已知A=x2+2x+1x2-1
4、-xx-1.
(1)化簡A;
(2)當x滿足不等式組x-1≥0,x-3<0且為整數(shù)時,求A的值.
13.化簡:aa2-4a+2a2-3a-12-a,并求值.其中a與2,3構(gòu)成△ABC的三邊長,且a為整數(shù).
拓展提升
14.[xx達州] 化簡代數(shù)式:( 3xx-1 - xx+1)xx2-1,再從不等式組x-2(x-1)≥1?、?6x+10>3x+1?、诘慕饧腥∫粋€合適的整數(shù)值代入,求出代數(shù)式的值.
15.對于任意的實數(shù)x,記f(x)=2x2x+1.
例如:f(1)=2121+1=23,f(-2)=2-22-2+1=15.
(1)求f(2),f(-3)的值;
(2)試猜想f(x
5、)+f(-x)的值,并說明理由;
(3)計算:f(-xx)+f(-xx)+…+f(-1)+f(0)+f(1)+…+f(xx)+f(xx).
參考答案
1.B [解析] 由等式x-3x+1=x-3x+1成立,可得x-3≥0,x+1>0,解得x≥3.故選B.
2.A [解析] 根據(jù)整式的運算法則及分式的基本性質(zhì)化簡,原式=x2+y2+2xy-x2-y2+2xy4xy=4xy4xy=1.
3.D [解析] 1x-1y=3,y-x=3xy,∴x-y=-3xy,
∴原式=2(x-y)+3xy(x-y)-xy=-6xy+3xy-3xy-xy=-3xy-4xy=34.
4.B 5.-1 6.2
6、n+1n2+1 7.-a-1
8.15ba-b 9.n+12n
10.解:(1)4a+4b5ab15a2ba2-b2=4(a+b)5ab15a2b(a+b)(a-b)=12aa-b.
(2)原式=x+1-1x+1x2-1x=xx+1(x+1)(x-1)x=x-1.
當x=2+1時,原式=2+1-1=2.
11.解:( 1a-b-ba2-b2)a2-aba2-2ab+b2
=a+b(a+b)(a-b)-b(a+b)(a-b)(a-b)2a(a-b)
=a+b-b(a+b)(a-b)a-ba
=aa+b1a
=1a+b.
由于a,b滿足a+b-12=0,
所以a+b=12,
7、
因此原式化簡后的式子:1a+b=112=2.
12.解:(1)A=(x+1)2(x+1)(x-1)-xx-1=x+1x-1-xx-1=x+1-xx-1=1x-1.
(2)解不等式組,得1≤x<3.
∵x為整數(shù),∴x=1或x=2.
∵A=1x-1,∴x≠1.
當x=2時,A=1x-1=12-1=1.
13.解:原式=a(a+2)(a-2)a+2a(a-3)+1a-2
=1(a-2)(a-3)+a-3(a-2)(a-3)
=a-2(a-2)(a-3)
=1a-3.
∵a與2,3構(gòu)成△ABC的三邊長,∴3-2
8、=2時,分母2-a=0,舍去;當a=3時,分母a-3=0,舍去;故a的值只能為4.
當a=4時,原式=14-3=1.
14.解:解不等式①,得x≤1,解不等式②,得x>-3,
∴不等式組x-2(x-1)≥1?、?6x+10>3x+1 ②的解集為-3