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受欢迎的初等代数 >

(32^2)+(5^2\div 5^2)+[4(2^4)]

解答

(3 2^{2}) + (5^{2} \div 5^{2}) + [4 (2^{4})]

解答

1089

求解步骤

(3 2^{2}) + (5^{2} \div 5^{2}) + [4 (2^{4})]

遵循 PEMDAS 运算顺序

计算括号内的值

(3 2^{2}) : 1024

3 2^{2}

3 2^{2} = 1024

= 1024

= 1024 + (5^{2} \div 5^{2}) + [4 (2^{4})]

计算括号内的值

(5^{2} \div 5^{2}) : 1

5^{2} \div 5^{2}

计算幂值

5^{2} : 25

5^{2}

5^{2} = 25

= 25

= 25 \div 5^{2}

计算幂值

5^{2} : 25

5^{2}

5^{2} = 25

= 25

= 25 \div 25

乘除（左右）

25 \div 25 : 1

25 \div 25

25 \div 25 = 1

= 1

= 1

= 1024 + 1 + [4 (2^{4})]

计算括号内的值

[4 (2^{4})] : 64

4 (2^{4})

计算括号内的值

(2^{4}) : 16

2^{4}

2^{4} = 16

= 16

= 4 \cdot 16

乘除（左右）

4 \cdot 16 : 64

4 \cdot 16

4 \cdot 16 = 64

= 64

= 64

= 1024 + 1 + 64

加减（左右）

1024 + 1 + 64 : 1089

1024 + 1 + 64

1024 + 1 = 1025

= 1025 + 64

1025 + 64 = 1089

= 1089

= 1089

流行的例子

3 \cdot 5 \cdot 4^{2}

(2+6× 3)\div 2+13

(2 + 6 \times 3) \div 2 + 13

1-4× (3-6\div 2)-5× 3

1 - 4 \times (3 - 6 \div 2) - 5 \times 3

1^1+2^2+3^3+4^4

1^{1} + 2^{2} + 3^{3} + 4^{4}

(- 4 - 2)^{2}