word格式-可编辑-感谢下载支持七年级数学上......一元一次方程练习题.........一.选择....题.1..在..a.-.(.b.-.c.).=.a.-.b.+.c.,.4.+.x.=.9.,.C.=.2..r.,.3.x.+.2.y.中等式的个数为.......()....(A)1....个.(B)2....个.(C)3....个.(D)4....个.2..在方程....6.x.+.1.=.1.,.2x23,7.x.-.1.=.x.-.1.,.5.x.=.2.-.x.中解为...13的方程个数是......()....(A)1....个.(B)2....个.(C)3....个.(D)4....个.3..根据等式性质.......5.=.3.x.-.2.可变形为....()....(A)...-.3.x.=.2.-.5.(B)...-.3.x.=-..2.+.5.(C)5....-.2.=.3.x.(D)5....+.2.=.3.x.4..下列方程中,解是.........x.=.4.的是..()....(A)2....x.+.4.=.9.(B)...32x23x4(C)...-.3.x.-.7.=.5.(D)5....-.3.x.=.2(1...-.x.).5..已知关于.....y.的方程...y.+.3.m.=.24..与.y.+.4.=.1.的解相同,则......m.的值是...()....(A)9....(B)...-.9.(C)7....(D)...-.8.6..方程...14x13正确的解是.....()....(A)...x.=.12..;.(B)...x112;.(C)...x43;.(D)...x347..将..3(..x.-.1)..-.2(..x.-.3)..=.5(1...-.x.).去括号得....()..(A)3....x.-.1.-.2.x.-.3.=.5.-.x.(B)3....x.-.1.-.2.x.+.3.=.5.-.x.(C)3....x.-.3.-.2.x.-.6.=.5.-.5.x.(D)3....x.-.3.-.2.x.+.6.=.5.-.5.x.8..已知关于.....x.的方程...(.a.+.1)..x.+.(4..a.-.1)..=.0.的解为-....2.,则..a.的值等于....()....(A)...-.2.(B)0....(C)...23(D)...329...已知..y.=.1.是方程...213(my)2y的解,...关于..x.的方程...m.(.x.-.3)..-.2.=.m.(2..x.-.5)..的解是...().(A)...x.=.10..(B)...x.=.0.(C)...x43(D)...x3410...方程...x315x16的解为...()(A).....73;.(B)...53;.(C)...353;.(D)...37311...若关于....x.的方程...2xa24(x1)的解为...x.=.3.,则..a.的值为...()....(A)2....(B)22.....(C)10.....(D)...-.2.12...方程...xx125的解为...()....(A)...-.9.;.(B)3....;.(C)...-.3.;.(D)9.....word格式-可编辑-感谢下载支持13.方程.....35x7x17().,去分母,得........24(A)3-2(5x+7)=-(x+17)....................(C)12-2(5x+7)=-(x+17).....................14.将....0.2(A)...2xx(B)12-2(5x+7)=-x+17...................(D)12-10x+14=-(x+17)....................0.50.01x().1的分母化为整数,得............0.03x0.50.01x130.50.01x1003(B)5x...(D)5x...50x100350x13(C)...2015.方程2xa1与方程3x12x2的解相同,则a的值为()A.-5B.-3C.3D.516.一商店把彩电按标价的九折出售,仍可获利%,若该彩电的进价是元,则彩电......................20............2400.........的标价为().(A)3200元(B)3429元(C)2667元(D)3168元.......................................17.某种手机卡的市话费上次已按原收费标准降低...