题目内容
解方程:
(1)
-
=1
(2)1-
=
(3)
-
=1
(4)
y+
=
y-
(5)
-
-
+3=0
(6)
=
-
.
(1)
| 2x+1 |
| 3 |
| 5x-1 |
| 6 |
(2)1-
| x+2 |
| 3 |
| x-1 |
| 2 |
(3)
| x-3 |
| 2 |
| 4x+1 |
| 5 |
(4)
| 11 |
| 9 |
| 2 |
| 7 |
| 2 |
| 9 |
| 5 |
| 7 |
(5)
| x-2 |
| 5 |
| x+3 |
| 10 |
| 2x-5 |
| 3 |
(6)
| 5x+1 |
| 6 |
| 9x+1 |
| 8 |
| 1-x |
| 3 |
分析:各方程去分母后,去括号,移项合并,将x系数化为1,即可求出解.
解答:解:(1)去分母得:2(2x+1)-(5x-1)=6,
去括号得:4x+2-5x+1=6,
移项合并得:-x=3,
解得:x=-3;
(2)去分母得:6-2(x+2)=3(x-1),
去括号得:6-2x-4=3x-3,
移项合并得:5x=5,
解得:x=1;
(3)去分母得:5(x-3)-2(4x+1)=10,
去括号后得:5x-15-8x-2=10,
移项合并得:-3x=27,
解得:x=-9;
(4)去分母得:77y+18=14y-45,
移项合并得:63y=-63,
解得:y=-1;
(5)去分母得:6(x-2)-3(x+3)-10(2x-5)+90=0,
去括号得:6x-12-3x-9-20x+50+90=0,
移项合并得:-17x=119,
解得:x=-7;
(6)去分母得:4(5x+1)=3(9x+1)-8(1-x),
去括号得:20x+4=27x+3-8+8x,
移项合并得:15x=9,
解得:x=
.
去括号得:4x+2-5x+1=6,
移项合并得:-x=3,
解得:x=-3;
(2)去分母得:6-2(x+2)=3(x-1),
去括号得:6-2x-4=3x-3,
移项合并得:5x=5,
解得:x=1;
(3)去分母得:5(x-3)-2(4x+1)=10,
去括号后得:5x-15-8x-2=10,
移项合并得:-3x=27,
解得:x=-9;
(4)去分母得:77y+18=14y-45,
移项合并得:63y=-63,
解得:y=-1;
(5)去分母得:6(x-2)-3(x+3)-10(2x-5)+90=0,
去括号得:6x-12-3x-9-20x+50+90=0,
移项合并得:-17x=119,
解得:x=-7;
(6)去分母得:4(5x+1)=3(9x+1)-8(1-x),
去括号得:20x+4=27x+3-8+8x,
移项合并得:15x=9,
解得:x=
| 3 |
| 5 |
点评: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.
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