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受欢迎的初等代数 >

5-37\div 12+8\div 93+3-11

解答

5 - 3 \cdot 7 \div 12 + 8 \div 9 \cdot 3 + 3 - 11

解答

- \frac{25}{12}

+1

十进制

- 2.08333 \dots

求解步骤

5 - 3 \cdot 7 \div 12 + 8 \div 9 \cdot 3 + 3 - 11

遵循 PEMDAS 运算顺序

乘除（左右）

3 \cdot 7 \div 12 : \frac{21}{12}

3 \cdot 7 \div 12

3 \cdot 7 = 21

= 21 \div 12

21 \div 12 = \frac{21}{12}

= \frac{21}{12}

= 5 - \frac{21}{12} + 8 \div 9 \cdot 3 + 3 - 11

乘除（左右）

8 \div 9 \cdot 3 : \frac{8}{3}

8 \div 9 \cdot 3

8 \div 9 = \frac{8}{9}

= \frac{8}{9} \cdot 3

\frac{8}{9} \cdot 3 = \frac{8}{3}

\frac{8}{9} \cdot 3

将项转换为分式:

3 = \frac{3}{1}

= \frac{8}{9} \cdot \frac{3}{1}

使用分式法则:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{8 \cdot 3}{9 \cdot 1}

\frac{8 \cdot 3}{9 \cdot 1} = \frac{8}{3}

\frac{8 \cdot 3}{9 \cdot 1}

数字相乘:

9 \cdot 1 = 9

= \frac{8 \cdot 3}{9}

因式分解数字:

9 = 3 \cdot 3

= \frac{8 \cdot 3}{3 \cdot 3}

约分:

3

= \frac{8}{3}

= \frac{8}{3}

= \frac{8}{3}

= 5 - \frac{21}{12} + \frac{8}{3} + 3 - 11

加减（左右）

5 - \frac{21}{12} + \frac{8}{3} + 3 - 11 : - \frac{25}{12}

5 - \frac{21}{12} + \frac{8}{3} + 3 - 11

5 - \frac{21}{12} = \frac{13}{4}

5 - \frac{21}{12}

消掉

\frac{21}{12} : \frac{7}{4}

\frac{21}{12}

约分:

3

= \frac{7}{4}

= 5 - \frac{7}{4}

将项转换为分式:

5 = \frac{5 \cdot 4}{4}

= \frac{5 \cdot 4}{4} - \frac{7}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 \cdot 4 - 7}{4}

5 \cdot 4 - 7 = 13

5 \cdot 4 - 7

数字相乘:

5 \cdot 4 = 20

= 20 - 7

数字相减:

20 - 7 = 13

= 13

= \frac{13}{4}

= \frac{13}{4} + \frac{8}{3} + 3 - 11

\frac{13}{4} + \frac{8}{3} = \frac{71}{12}

\frac{13}{4} + \frac{8}{3}

4, 3

的最小公倍数:

12

4, 3

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

将每个因子乘以它在

4

或

3

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{13}{4} :

将分母和分子乘以

3

\frac{13}{4} = \frac{13 \cdot 3}{4 \cdot 3} = \frac{39}{12}

对于

\frac{8}{3} :

将分母和分子乘以

4

\frac{8}{3} = \frac{8 \cdot 4}{3 \cdot 4} = \frac{32}{12}

= \frac{39}{12} + \frac{32}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{39 + 32}{12}

数字相加:

39 + 32 = 71

= \frac{71}{12}

= \frac{71}{12} + 3 - 11

\frac{71}{12} + 3 = \frac{107}{12}

\frac{71}{12} + 3

将项转换为分式:

3 = \frac{3 \cdot 12}{12}

= \frac{3 \cdot 12}{12} + \frac{71}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{3 \cdot 12 + 71}{12}

3 \cdot 12 + 71 = 107

3 \cdot 12 + 71

数字相乘:

3 \cdot 12 = 36

= 36 + 71

数字相加:

36 + 71 = 107

= 107

= \frac{107}{12}

= \frac{107}{12} - 11

\frac{107}{12} - 11 = - \frac{25}{12}

\frac{107}{12} - 11

将项转换为分式:

11 = \frac{11 \cdot 12}{12}

= - \frac{11 \cdot 12}{12} + \frac{107}{12}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 11 \cdot 12 + 107}{12}

- 11 \cdot 12 + 107 = - 25

- 11 \cdot 12 + 107

数字相乘:

11 \cdot 12 = 132

= - 132 + 107

数字相加/相减:

- 132 + 107 = - 25

= - 25

= \frac{- 25}{12}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{25}{12}

= - \frac{25}{12}

= - \frac{25}{12}

流行的例子

20 + 5 (38)

2 (3 + \frac{7}{8}) - \frac{1}{3}

(-2)*(-2)*(-2)*(-2)

(- 2) \cdot (- 2) \cdot (- 2) \cdot (- 2)

6^{2} - 5 (6) - 3

(17 \cdot 16)^{4}