试题

题目:
解下列方程
(1)4y-下(20-y)=5y-6
(2)
x-1
2
-
2x-1
6
=
x+1
-1

答案
解:(1)去括号得:4多-60+3多=5多-6,
移项得:4多+3多-5多=-6+60,
合并得:2多=54,
解得:多=27;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;
(2)去分母得:3(x-1)-(2x-1)=2x+2-6,
去括号得:3x-3-2x+1=2x+2-6,
移项得:3x-2x-2x=2-6-1+3,
合并得:-x=-2,
解得:x=2.
解:(1)去括号得:4多-60+3多=5多-6,
移项得:4多+3多-5多=-6+60,
合并得:2多=54,
解得:多=27;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;&nbs7;
(2)去分母得:3(x-1)-(2x-1)=2x+2-6,
去括号得:3x-3-2x+1=2x+2-6,
移项得:3x-2x-2x=2-6-1+3,
合并得:-x=-2,
解得:x=2.
考点梳理
解一元一次方程.
(1)利用去括号法则去括号后,移项合并,将x系数化为1,即可求出解;
(2)两边乘以6去分母后,去括号,移项合并,将x系数化为1,即可求出解.
此题考查了解一元一次方程,其步骤为:去分母,去括号,移项合并,将未知数系数化为1,求出解.
计算题.
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