1 - 2 mũ 1 + 2 mũ 2 - 2 mũ 3 + 2 mũ 4 - 2 mũ 5 + ... - 2 mũ 19 + 2 mũ 20
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GH
Gia Hân
VIP
10 tháng 8 2023
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
22 tháng 12 2017
a,\(5^3.2-100:4+2^3.5\)
= 125 . 2 - 25 + 8 . 5
= 250 - 25 + 40
= 265
b, \(6^2:9+50.2-3^3.3\)
= 36 : 9 + 100 - 27 . 3
= 4 + 100 - 81
= 23
ND
Nguyễn Đức Trí
VIP
27 tháng 8 2023
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
ND
Nguyễn Đức Trí
VIP
27 tháng 8 2023
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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
ND
Nguyễn Đức Trí
VIP
15 tháng 8 2023
a) \(9^{21}.9^{33}=9^{21+33}=9^{54}\)
b) \(19^{11}.19.19=19^{11+1+1}=19^{13}\)
c) \(25^2.5^2.125=5^4.5^2.5^3=5^{4+2+3}=5^9\)
d) \(t^{2021}.t^2.\left(t^2\right)^2=t^{2021}.t^2.t^4=t^{2021+2+4}=t^{2027}\)
e) \(123^{14}:123^{13}=123^{14-13}=123\)
f) \(64^2:8^3=\left(8^2\right)^2:8^3=8^4:8^3=8^{4-3}=8=2^3\)
g) \(6^{10}:6^3:36=6^{10}:6^3:6^2=6^{10-3-2}=6^5\)
h) \(m^{20}:m^{10}.m^{10}=m^{20-10+10}=m^{20}\)
2 tháng 8 2021
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
2 tháng 8 2021
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
19 tháng 9 2018
- 22.32.5:22.3-32=3.5-32=15-9=6
- 2.52-22.32:32=2.(52-2)=2.(25-2)=46
19 tháng 9 2018
3. 33.19-33.12=33.(19-12)=33.7=189
4. 3.52-16:22=3.52-24:22=3.25-4=75-4=71
Đặt A=1-2^1+2^2-2^3+...+2^20
=>2A=2-2^2+2^3-2^4+...+2^21
=>2A+A=(2-2^2+2^3-2^4+...+2^21)+(1-2^1+2^2-2^3+...+2^20)
<=>3A=2^21-1
=>A=\(\frac{2^{21}-1}{3}\)
\(A=1-2^1+2^2-2^3+...-2^{19}+2^{20}\)
\(A=\left(1+2^2+...+2^{10}\right)-\left(2^1+2^3+...+2^{19}\right)\)
Đặt \(B=1+2^2+...+2^{10}\)
\(2^2B=2^2+2^4+...+2^{12}\)
\(4B-B=\left(2^2+2^4+...+2^{12}\right)-\left(1+2^2+...+2^{10}\right)\)
\(3B=2^{12}-1\)
\(B=\frac{2^{12}-1}{3}\)
Đặt \(C=2^1+2^3+...+2^{19}\)
\(2^2C=2^3+2^5+...+2^{21}\)
\(4C-C=\left(2^3+2^5+...+2^{21}\right)-\left(2^1+2^3+...+2^{19}\right)\)
\(3C=2^{21}-2\)
\(C=\frac{2^{21}-2}{3}\)
Ta có :
\(A=B-C\)
hay \(A=\frac{2^{12}-1}{3}-\frac{2^{21}-2}{3}\)
\(A=\frac{2^{12}-1-2^{21}+2}{3}\)
\(A=\frac{-2093055}{3}\)
\(A=-697685\)