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85+(-35)+(-12)+(-35)=50+(-12)+(-35)=38+(-35)=3
53.42+85.53-57:54=125.16+85.125-53=125.16+85.125-125
=125.16+85.125-125.1=(16+85-1).125=100.125=12500
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5^2\cdot2\)
\(=5\cdot\left[\left(85-5\right):8+90\right]-25\cdot2\)
\(=5\cdot\left(10+90\right)-50\)
=450
a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
\(5^{12}.7-5^{11}.10\)
\(=5^{11}.\left(5.7-10\right)\)
\(=5^{11}.25\)
\(=5^{11}.5^2\)
\(=5^{13}\)
\(2^{20}.15+2^{20}.85\)
\(=2^{20}.5\left(3+17\right)\)
\(=2^{20}.100\)
\(=104857600\)
\(125^3:25^4\)
\(=\left(5^3\right)^3:\left(5^2\right)^4\)
\(=5^9:5^8\)
\(=5\)
\(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:3\right)^4-\left(32:16\right)^{12}\)
\(=8^4-2^{12}\)
\(=0\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
1. 53 = 5.5.5 = 125
2. 27 = 2.2.2.2.2.2.2 = 128
3. 44 = 4.4.4.4 = 256
4. 73 = 7.7.7 = 343
6. 35 = 243
7. 26 = 64
8. 34 = 81
9. 83 = 512
11. 132 = 169
12. 112 = 121
13. 142 = 196
14. 152 = 225
16. 172 = 289
17. 182 = 324
18. 192 = 361
19. 202 = 400
21. 104 = 10000
22. 105 = 100000
23. 106 = 1000000
24. 107 = 10000000
Giải: (^ là mũ)
a) 3^8 : 3^4 + 2^2 . 2^3
= 6561 : 81 + 4 . 8
= 81 + 32
= 113
Mấy câu sau bn làm tương tự nhé!
b: \(=2^{12}\cdot3^4\cdot\dfrac{3^{10}}{3^{12}\cdot2^{12}}=3^2=9\)
c: \(=\dfrac{7^2\cdot3^2\cdot7\cdot2\cdot5^3}{5^3\cdot7^3\cdot2\cdot3}=\dfrac{3^2}{3}=3\)
d: \(=\dfrac{5^3\cdot3^6\cdot5^4\cdot2^8\cdot3^{16}\cdot2^2}{\left(2^2\cdot3^2\cdot5\right)^5}=\dfrac{5^{10}\cdot3^{22}\cdot2^{10}}{2^{10}\cdot3^{10}\cdot5^5}=3^{12}\cdot5^5\)
\(2^3-5^3:5^2+12\cdot2^2=8-5+48=51\\ 5\left[\left(85-35:7\right):8+90\right]-5\cdot2^2\\ =5\left(80:8+90\right)-20=5\cdot100-20=480\)