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受欢迎的初等代数 >

(1(6))\div (1(2))(1(1))+(1(2))

解答

(1 (6)) \div (1 (2)) (1 (1)) + (1 (2))

解答

5

求解步骤

(1 \cdot (6)) \div (1 \cdot (2)) (1 \cdot (1)) + (1 \cdot (2))

遵循 PEMDAS 运算顺序

计算括号内的值

(1 (6)) : 6

1 \cdot (6)

1 (6) = 6

= 6

= 6 \div (1 \cdot (2)) (1 \cdot (1)) + (1 \cdot (2))

计算括号内的值

(1 (2)) : 2

1 \cdot (2)

1 (2) = 2

= 2

= 6 \div 2 (1 \cdot (1)) + (1 \cdot (2))

计算括号内的值

(1 (1)) : 1

1 \cdot (1)

1 (1) = 1

= 1

= 6 \div 2 \cdot 1 + (1 \cdot (2))

计算括号内的值

(1 (2)) : 2

1 \cdot (2)

1 (2) = 2

= 2

= 6 \div 2 \cdot 1 + 2

乘除（左右）

6 \div 2 \cdot 1 : 3

6 \div 2 \cdot 1

6 \div 2 = 3

= 3 \cdot 1

3 \cdot 1 = 3

= 3

= 3 + 2

加减（左右）

3 + 2 : 5

3 + 2

3 + 2 = 5

= 5

= 5

流行的例子

6+12\div (12-6)

6 + 12 \div (12 - 6)

(-1+2)(-1^2-2(-1)+4)

(- 1 + 2) (- 1^{2} - 2 (- 1) + 4)

- 7 + 19 - 23 - 14 + 5

(2\div 2/3)-(2/3 \div 2)

(2 \div \frac{2}{3}) - (\frac{2}{3} \div 2)

- \frac{1}{3} + 3 \div 2