表示 2.684684684… 作为有理数
有理数是可以表示或写成 m/n 形式的数字,其中 m 和 n 是整数,n ≠ 0。由于数字的基本结构 m/n,大多数人难以区分分数和有理数形式。当一个有理数被拆分时,结果是一个十进制值,它可能是结束的,也可能是重复的。有理数的例子有 11、-11、5、-5、9 等等,它们可以用分数形式写成 11/1、-5/1 和 7/1。
有理数是一种实数,公式为 m/n,其中 n≠0。有理数除以十进制数,可以结束也可以重复。
十进制数到有理数的转换
这里给出了将十进制数转换为有理数的步骤,
- 第 1 步:识别重复小数并将其等于 x。
- 第 2 步:通过从重复数字顶部删除条并列出重复数字至少两次,以十进制形式写出来。
For Example, write x = 0.3 bar as x = 0.333… and x = 0.33 bar as x = 0.333333…
- 第 3 步:检查有条的位数。
- 第 4 步:如果有重复小数的数字有 1 位重复,那么我们将它乘以 10,如果它有两位重复,那么它将乘以 100,并且三位重复乘以 1000,以此类推在。
- 第五步:然后从第四步得到的方程中减去第二步得到的方程。
- 第 6 步:无论剩下什么,将等式两边除以 x 系数。
- 第7步:最后,写出最简单的有理数。
表示 2.684684684… 作为有理数
解决方案:
Given: 2.684684684.. or
Lets assume x = 2.684684684… ⇢ (1)
And, there are three digits after decimal which are repeating,
So, multiply equation (1) both sides by 1000,
So, 1000 x = 2684.684684 ⇢ (2)
Now subtract equation (1) from equation (2)
1000x – x = 2684.684684.. – 2.684684..
999x = 2682
x = 2682/999
= 894/ 333
= 298/111
2.684684684.. can be expressed 298/111 as rational number
类似问题
问题 1:将 356.68686868... 表示为 p/q 形式的有理数,其中 p 和 q 没有公因数。
解决方案:
Given: 356.68686868… or
Lets assume x = 356.68686868… ⇢ (1)
And, there are two digits after decimal which are repeating,
So, multiply equation (1) both sides by 100,
So 100 x = ⇢ (2)
Now subtract equation (1) from equation (2)
100x – x =
99x = 35312
x = 35312/99
356.68686868… can be expressed 35312/99 as rational number.
问题 2:将 31.247247247... 表示为 p/q 形式的有理数,其中 p 和 q 没有公因数。
解决方案:
Given: 31.247247247 or
Let’s assume x = 31.247247247… ⇢ (1)
And, there are three digits after decimal which are repeating
So multiply equation (1) both sides by 1000
So, 1000x = ⇢ (2)
Now subtract equation (1) from equation (2)
1000x – x =
999x = 31216
9x = 31216/999
31.247247247… can be expressed 31216/999 as rational number
问题 3:将 105.357357357... 表示为一个有理数,形式为 p/q,其中 p 和 q 没有公因数。
解决方案:
Given: 105.357357357… or
Let’s assume x = 105.357357357… ⇢ 1
And, there are three digits after decimal which are repeating,
So multiply equation 1 both sides by 1000
So 1000 x = ⇢ (2)
Now subtract equation (1) from equation (2)
1000x – x =
999x = 105252
x = 105252/ 999
= 35084/333
105.357357357 can be expressed 35084/333 in form of p/q as rational number.
问题 4:将 14.777777... 表示为一个有理数,形式为 p/q,其中 p 和 q 没有公因数。
解决方案:
Given: 14.777777… or
Let’s assume x = 14.777777…. ⇢ (1)
And, there are two digits after decimal which are repeating, so multiply equation (1) both sides by 100,
So 100 x = ⇢ (2)
Now subtract equation (1) from equation (2)
100x – x =
99x = 1463
x = 1463 /99
14.777777…. can be expressed 1463/99 in form of p/q as rational number .
问题 5:将 157.927927927... 表示为一个有理数,形式为 p/q,其中 p 和 q 没有公因数。
解决方案:
Given: 157.927927927… or
Let’s assume x = 157.927927927… ⇢ 1
And, there are three digits after decimal which are repeating,
So multiply equation (1) both sides by 1000,
So 1000 x = ⇢ (2)
Now subtract equation (1) from equation (2)
1000x – x =
999x = 157770
x = 157770/999
157.927927927… can be expressed 157770/999 in form of p/q as rational number.
问题 6:将 2.252525... 表示为有理数,形式为 p/q,其中 p 和 q 没有公因数。
解决方案:
Given: 2.252525….
Let’s assume x = 2.252525…… ⇢ (1)
And, there are two digits after decimal which are repeating, so multiply equation (1) both sides by 100,
So 100 x = ⇢ (2)
Now subtract equation (1) from equation (2)
100x – x =
99x = 223
x = 223/99
2.252525…. can be expressed 223/99 in form of p/q as rational number.
问题 7:将 0.111111... 表示为有理数,形式为 p/q,其中 p 和 q 没有公因数。
解决方案:
Given: 0.111111…
Let’s assume x = 0.111111…. ⇢ (1)
And, there are two digits after decimal which are repeating, so multiply equation (1) both sides by 100,
So 100 x = ⇢ (2)
Now subtract equation (1) from equation (2)
100x – x =
99x = 11
x = 11/99
= 1/9
0.111111…. can be expressed 1/9 in form of p/q as rational number .