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。。。
2024-11-22 19:31:16
推荐回答(1个)
回答1:

(1)
2x+1/6-4x-1/8=1
2x - 4x + 1/24 = 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,方程无解。

若写成:
(x-1)/0.5-(x+2)/0.3=2
2x-2 - (10/3)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