解方程:(x^2+X+1)⼀(x^2+1)+(2x^2+x+2)⼀(x^2+x+1)=19⼀6

2025-02-23 12:29:48
推荐回答(5个)
回答1:

1、苹果5元1斤,梨8元1斤,各买X斤,用了78元,问都买了几斤?
解:5x+8x=78
13x=78
x=6
答:各买了6斤。

2、甲、乙两车间生产同一种机器,甲车间每天生产80台,乙车间每天生产120台,若甲先生产5天,问乙要生产多少天才能赶上甲的产量?
解:设X天才能赶上甲的产量
(120-80)X=80*5
40X=400
X=10
答:乙要生产10天才能赶上甲的产量。

3、某航空公司规定:乘坐飞机普通舱旅客一人最多可免费托运20千克行李,超过部分每千克按飞机票价的1.5%购买行李票。一名旅客托运了35千克行李,机票连同行李费共付1323元,求该旅客的机票票价?
解:设该机票价格为X元
则X+(35-20)*1.5%*X=1323
X+0.225X=1323
合并同类项1.225X=1323
方程两边同时除以1.225,得X=1080
答:该旅客的家票价格是1080元。

4、现对某商品降价10%促销,为了使销售总金不变,销售量要比原价销售时增加百分之几~还有一题:甲骑自行车从A地到B地,乙骑自行车从B地到A地,两人都匀速前进,已知两人在上午8时同时出发,上午10时,两人还相距36千米,到中午12时,两人又相距36千米.求A.B两地间路程?
商品降价10%促销后,价格变为原来的90%,总金额不变,那销售量就得是原来的
1/90%=10/9
销售量要比按原价销售时增加(10/9-1)*100%=11.1%

设甲的速度x千米/小时,乙的速度y千米/小时,AB路程为S
所以S=36+2*(x+y)=4*(x+y)-36
解方程组得:x+y=36千米/小时
S=108千米
答:AB两地的路程为108千米

上午10时两人还相距36千米,到中午12时两人又相距36千米,则2小时两人共行36*2=72千米.
所以AB两地间的路程为72+36=108千米.

5、课外活动中一些学生分组活动,原来每组8人,后来重新编组,每组12人,这样比原来减少了两组.问这些学生共有多少人?
8X=12(X-2)
解方程去

6、解方程:(2x+1)*(2x-3)=4*(x-2)*(x+5)-11
4x*x-4x-3=4(x*x+3x-10)-11
-4x-3=12x-51
16x=-48
x=-3

7、爷爷用讲故事的办法教孙子数学,爷爷说,抗日战争时期,在某山区我侦察兵发现一对敌军和一群狗,它们排成一队走;人头狗头七十个,共用2百条腿走,请你算算多少只狗,多少个敌人?
设有x个人,y条狗
x+y=70
2x+4y=200
解得
x=40
y=30
有40个人,30条狗

8、已知一个三角形两边的长分别是a,b,且a〉b,那么这个三角形的周长l的取值范围是? 已知a+b=1,ab=-1,设S1=a+b,S2=a的平方+b的平方,Sn=a的n次方+b的n次方,计算S1,S2,S3,S4 写出Sn-2,Sn-1,Sn三者之间的关系式(注意n-2这些数是在S底下) 根据以上结论,a的7次方+b的七次方是?
问题1:因为第三条边的范围是大于a、b的差,小于a、b之和,根据S=a+b+c,显而易见S大于(a+b),再根据c<(a+b)可得,所以周长S范围是a+b问题2:不敢涉嫌抄袭,跟楼上意见一样。
回答者:天鹅想吃癞蛤蟆 - 魔法学徒 一级 6-22 08:01

解方程组:{ a+b=1 a*b=-1 }
得 a=-0.618033988749989 b=1.61803398874999
或 a=1.61803398874999 b=-0.618033988749989
问题2:Sn=a^n+b^n=(a+b)(a^n-1 + b^n-1)-[b*a^(n-1)+a*b^(n-1)]
=(a+b)*Sn-1-ab(a^(n-2)+b^(n-2))
=(a+b)*Sn-1-ab*Sn-2
=Sn-1 + Sn-2
所以
S1=1
S2=a^2+b^2=(a+b)^2-2ab=1+2=3
S3=1+3=4
S4=3+4=7
S5=4+7=11
S6=7+11=18
S7=11+18=29

9、甲、乙两人练习短跑,甲每秒跑7米,乙每秒跑6.5米,如果甲让乙先跑1秒,甲经过几秒钟可以追上乙?
解:设x秒后甲追上乙。
7x=6.5(x+1)
7x=6.5x+6.5
0.5x=6.5
x=13
答:13秒后甲追上乙。

10、某班学生每小时4KM的速度从学校不行到农场参加活动,走了1.5小时后,小王奉命去取东西,他每小时行6KM的速度取了之后立即又以同样的速度追赶队伍,截获距离2KM的地方追上了队伍,求学校和农场的距离?
1.设小王返回时走了x小时。
1.5*4+2=6x
x=4/3
4/3*6+2=10km
答:学校距离农场10千米。

回答2:

原式<=>(x^2+x+1)/(x^2+1)+(x^2+1)/(x^2+x+1)=13/6 令t=(x^2+x+1)/(x^2+1) 则3/2>=t>=1/2 所以 原式<=>t+1/t=13/6 解得t=2/3或3/2 即(x^2+x+1)/(x^2+ 1)=2/3或3/2 解得x=1

回答3:

X=1.26

回答4:

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回答5:

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