(2)(2
﹣)(﹣2)﹣.
【分析】(1)分别进行平方差公式和完全平方公式的运算,然后合并; (2)先进行二次根式的乘法和二次根式的化简,然后合并. 【解答】解:(1)原式=3+5﹣2+2
=10﹣2
;
(2)原式=6﹣4﹣+4﹣3+2
=9﹣5.
19.解方程组: (1)
(2)
.
【分析】(1)方程组利用加减消元法求出解即可; (2)方程组整理后,利用加减消元法求出解即可. 【解答】解:(1)
,
①+②得:14x=﹣7,即x=﹣, 把x=﹣代入①得:y=﹣3, 则方程组的解为;
(2)方程组整理得:
,
①+②×5得:14y=28,即y=2, 把y=2代入②得:x=2, 则方程组的解为.
20.某商场按定价销售某种商品时,每件商品可以获利50元.已知按定价的八
第13页(共17页)
折销售该商品3件与将定价降低20元销售该商品2件所获得的利润相等.请求出该商品的进价和定价分别是多少?
【分析】设该商品的进价为x元,则定价为(x+50)元,根据按定价的八折销售该商品3件与将定价降低20元销售该商品2件所获得的利润相等,列方程求解.
【解答】解:设该商品的进价为x元,则定价为(x+50)元, 由题意得,3×[(x+50)×0.8﹣x]=2×30, 解得:x=100.
则商品的定价为100+50=150(元). 答:商品的进价为100元,定价为150元.
21.列方程组解应用题:
甲乙两人从相距36千米的两地相向而行.如果甲比乙先走2小时,那么在乙出发后3小时相遇;如果乙比甲先走2小时,那么在甲出发后2.5小时相遇.甲、乙两人每小时各走多少千米?
【分析】设甲,乙速度分别为x,y千米/时,根据甲乙两人从相距36千米的两地相向而行.如果甲比乙先走2小时,那么在乙出发后3时相遇;如果乙比甲先走2小时,那么在甲出发后2.5时相遇可列方程求解. 【解答】解:设甲,乙速度分别为x,y千米/时,
,
,
甲的速度是3.6千米每小时,乙的速度是6千米每小时.
22.已知方程组
了方成组中的m.得到
,由于甲看错了方程组中的n,解得,乙看错
,请你帮助他们按正确的m、n解原方程组.
【分析】把x=1,y=3代入方程组第一个方程求出m的值,把x=11,y=2代入第
第14页(共17页)
二个方程求出n的值,确定出原方程组的解即可. 【解答】解:把x=1,y=3代入得:m﹣9=﹣3,即m=6; 把x=11,y=2代入得:55﹣2n=﹣35,即n=45, 方程组为
,
,
①﹣②×2得:17y=13,即y=把y=
代入②得:x=﹣
,
则方程组的解为.
23.A、B两地相距100千米,甲、乙两人骑车同时分别从A,B两地相向而行.假设他们都保持匀速行使,则他们各自到A地的距离s(千米)都是骑车时间t(时)的一次函数.
(1)甲的速度为 15km/h ,乙的速度为 20km/h ; (2)求出:l1和l2的关系式; (3)问经过多长时间两人相遇.
【分析】(1)利用图象上点的坐标得出甲、乙的速度即可; (2)利用待定系数法求出一次函数解析式即可; (3)利用两函数相等进而求出相遇的时间.
【解答】解:(1)如图所示:甲的速度为:30÷2=15(km/h), 乙的速度为:(100﹣80)÷1=20(km/h), 故答案为:15km/h;20km/h;
第15页(共17页)
(2)设l1的关系式为:s1=kt,则30=k×2,解得:k=15,故s1=15t; 设s2=at+b,将(0,100),(1,80),则
,
解得:
,
故l2的关系式为s2=﹣20t+100;
(3)当s1=s2,则15t=﹣20t+100, 解得:t=故经过
.
小时两人相遇.
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