\(10-\left\{\left[3^x+1:\left(3+17\right):10+3\cdot2^4\right]:10\right\}=5\)

\(\Rightarrow\left(3^x+1:20:10+3\cdot16\right):10=10-5\)

\(\Rightarrow\left(3^x+\dfrac{1}{200}+48\right):10=5\)

\(\Rightarrow3^x+\dfrac{9601}{200}=5\cdot10\)

\(\Rightarrow3^x+\dfrac{9601}{200}=50\)

\(\Rightarrow3^x=50-\dfrac{9601}{200}\)

\(\Rightarrow3^x=\dfrac{399}{200}\)

Xem lại đề 

a: =30-22=8

b: =10*(23+17)=10*40=400

c: =21*3-7*(-14)

=63+98=161

d: =-20-[10*10*5^2+16]

=-20-100*25-16

=-36-2500

=-2536

chịu

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Chịu thôi🙂

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

bạn làm như bạn gia hân là đúng nhé

giải giúp mình bài 2 trên với ạ

2.4^x - 1 = 128

=> 4^x - 1 = 64

=> x - 1 = 3

=> x = 4

b, 17.4^2 + 83.16 

= 17.16 + 83.16

= 16(17 + 83)

= 16.100

= 1600

c, a^5 . 10^10.a^15

= a^20.10^10

d, 3{65 - [5(4 + 2.3) - 15] : 7}

= 3{65 - [5(4 + 6) - 15] : 7}

= 3{65 - [5.10 - 15] : 7}

= 3{65 - [50 - 15] : 7}

= 3{65 - 35 : 7}

= 3{65 - 5}

= 3.60

= 180

\(4^2\cdot120-4^3\cdot17+4^2\cdot34\)

\(=4^2\left(120-4\cdot17+34\right)\)

\(=16\left(120-68+34\right)\)

\(=16\cdot86=1088\)

\(7^3\cdot9+3^2\cdot7^4-45\cdot539\)

\(=7^3\cdot9+9\cdot7^4-45\cdot7^2\cdot11\)

\(=7^2\left(7\cdot9+9\cdot7^2-45\cdot11\right)\)

\(=49\left(63+9\cdot49-495\right)\)

\(=49\left(63+441-495\right)\)

\(=49\cdot9=441\)

\(5\cdot10^4+4\cdot10^3+6\cdot10^2+5\cdot10+9\)

\(=10\left(5\cdot10^3+4\cdot10^2+10\cdot6\right)+9\)

\(=10\left(5\cdot1000+4\cdot100+60\right)+9\)

\(=10\left(5000+400+60\right)+9\)

\(=10\cdot5460+9=54600+9=54609\)

\(A=\dfrac{10^{17}+3}{10^{17}+1}=1+\dfrac{2}{10^{17}+1}\\ B=\dfrac{10^{18}+1}{10^{18}-1}=1+\dfrac{2}{10^{18}-1}=1+\dfrac{2}{10^{17}+1+\left(9\cdot10^{17}-2\right)}\)

Ta có : \(9\cdot10^{17}-2>0\Rightarrow10^{17}+1+\left(9\cdot10^{17}-2\right)>10^{17}+1\\ \Rightarrow\dfrac{2}{10^{17}+1}>\dfrac{2}{10^{18}-1}\Rightarrow A>B\)

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)