2x+1⼀6-4x-1⼀8=1 x-1⼀0.5-x+2⼀0.3=2 3⼀4[3⼀4(12x+1⼀4)-8]=3⼀2x+1

怎么算呀?要过程哦。。。谢了。。。
2024-11-07 01:49:58
推荐回答(1个)
回答(1):

(1)
2x+1/6-4x-1/8=1
2x
-
4x
+
1/绝拿24
=
1
-2x
=
23/24
x
=
-23/48
若写成:
(2x+1)/6-(4x-1)/8=1
则方程两边都乘以24
(2x+1)*4
-
(4x-1)*3
=
24
8x
+
4
-
12x
+3
=
24
-4x
=
17
x
=
-17/4
(2)
x-1/0.5-x+2/0.3=2
因为左边x-x消去了未知数颤宏激,方程无解。
若写成:
(x-1)/0.5-(x+2)/0.3=2
2x-2
-
(10/茄袜3)x
-
20/3
=
2
(-4/3)x
=
32/3
x
=
-8
(3)
3/4[3/4(12x+1/4)-8]=3/2x+1
理解为:
(3/4)
*
[(3/4)
*
(12x+1/4)
-
8]
=
(3/2)x
+
1
(3/4)
*
[9x
+
3/16
-
8]
=
(3/2)x
+
1
(27/4)x
+
9/64
-
6
=
(3/2)x
+
1
(21/4)x
=
7
-
9/64
x
=
439/112