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受欢迎的初等代数 >

5 3/8+(6 3/4-4 1/8)

解答

5 \frac{3}{8} + (6 \frac{3}{4} - 4 \frac{1}{8})

解答

8

求解步骤

5 \frac{3}{8} + (6 \frac{3}{4} - 4 \frac{1}{8})

将带分数转换为假分式:

5 \frac{3}{8} = \frac{43}{8}

5 \frac{3}{8}

将带分数转换为假分式:

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

5 \frac{3}{8} = \frac{5 \cdot 8 + 3}{8}

= \frac{43}{8}

= \frac{43}{8} + (6 \frac{3}{4} - 4 \frac{1}{8})

将带分数转换为假分式:

6 \frac{3}{4} = \frac{27}{4}

6 \frac{3}{4}

将带分数转换为假分式:

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

6 \frac{3}{4} = \frac{6 \cdot 4 + 3}{4}

= \frac{27}{4}

= \frac{43}{8} + (\frac{27}{4} - 4 \frac{1}{8})

将带分数转换为假分式:

4 \frac{1}{8} = \frac{33}{8}

4 \frac{1}{8}

将带分数转换为假分式:

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

4 \frac{1}{8} = \frac{4 \cdot 8 + 1}{8}

= \frac{33}{8}

= \frac{43}{8} + (\frac{27}{4} - \frac{33}{8})

遵循 PEMDAS 运算顺序

计算括号内的值

(\frac{27}{4} - \frac{33}{8}) : \frac{21}{8}

\frac{27}{4} - \frac{33}{8}

\frac{27}{4} - \frac{33}{8} = \frac{21}{8}

\frac{27}{4} - \frac{33}{8}

4, 8

的最小公倍数:

8

4, 8

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

8

质因数分解:

2 \cdot 2 \cdot 2

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

将每个因子乘以它在

4

或

8

中出现的最多次数

= 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 = 8

= 8

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

8

对于

\frac{27}{4} :

将分母和分子乘以

2

\frac{27}{4} = \frac{27 \cdot 2}{4 \cdot 2} = \frac{54}{8}

= \frac{54}{8} - \frac{33}{8}

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{54 - 33}{8}

数字相减:

54 - 33 = 21

= \frac{21}{8}

= \frac{21}{8}

= \frac{43}{8} + \frac{21}{8}

加减（左右）

\frac{43}{8} + \frac{21}{8} : 8

\frac{43}{8} + \frac{21}{8}

\frac{43}{8} + \frac{21}{8} = 8

\frac{43}{8} + \frac{21}{8}

使用法则

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

= \frac{43 + 21}{8}

数字相加:

43 + 21 = 64

= \frac{64}{8}

数字相除:

\frac{64}{8} = 8

= 8

= 8

= 8

流行的例子

(42\div 2× 5)-(18-8)

(42 \div 2 \times 5) - (18 - 8)

8 \cdot 3^{2} - 5 \cdot 4 + 9

(- 3)^{2} - 3^{2} - 3^{0}

2 (0) + (1) - 4 (2)

1 4^{2} + 8^{2}