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500-{5.(409-(2³x3-21)²]-1724}
= 500-{5.(409-(8x3-21)²]-1724}
=500-{5.(409-(24-21)²]-1724}
=500-{5.(409-3²)-1724}
=500-{5.(409-9)-1724}
=500-{5.400-1724}
=500-{2000-1724}
=500-276
=224
Hok tốt!
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
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
\(7\cdot4^x=112\)
\(\Rightarrow4^x=\dfrac{112}{7}\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
_____
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=\dfrac{250}{2}\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
____
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+25\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left[400:\left(500-300\right)\right]+10\)
\(=12:\left(400:200\right)+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
\(7\cdot4^x=112\)
\(\Rightarrow4^x=112:7\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
\(---\)
\(2\cdot5^{x-3}=250\)
\(\Rightarrow5^{x-3}=250:2\)
\(\Rightarrow5^{x-3}=125\)
\(\Rightarrow5^{x-3}=5^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=3+3\)
\(\Rightarrow x=6\)
Vậy \(x=6.\)
\(---\)
\(12:\left\{400:\left[500-\left(5^3+5^2\cdot7\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-\left(125+175\right)\right]\right\}+10\)
\(=12:\left\{400:\left[500-300\right]\right\}+10\)
\(=12:\left\{400:200\right\}+10\)
\(=12:2+10\)
\(=6+10\)
\(=16\)
#\(Toru\)
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: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
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) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
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500 - { 5.[ 409 - (23 x 3 -21)2 ] + 103 } : 15
= 500 - { 5. [409 - (8x3 - 21)2] + 1000}:15
= 500 - { 5. [ 409 - (24 - 21)2] + 1000} : 15
= 500 - {5. (409 - 9) + 1000}:15
= 500 - { 5.400 + 1000}:15
= 500 - (2000+ 1000): 15
= 500 - 3000:15
= 500 - 200
= 300
300