\(|\)X + 9 \(|\)+ \(|\)6-x \(|\) + 450
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\(\frac{1}{2}\cdot4^n+4\cdot4^n=9\cdot2^{n+1}\)
\(\Rightarrow4^n\left(\frac{1}{2}+4\right)=9\cdot2^{n+1}\)
\(\Rightarrow2^{2n}\cdot\frac{9}{2}=9\cdot2^{n+1}\)
\(\Rightarrow2^{2n}=9\cdot2^{n+1}:\frac{9}{2}\)
\(\Rightarrow2^{2n}=9\cdot2^{n+1}\cdot\frac{2}{9}\)
\(\Rightarrow2^{2n}=2^{n+2}\)
\(\Rightarrow2n=n+2\)
\(\Rightarrow n=2\)
a, \(S=1+3+3^2+...+3^{2019}\)
\(3S=3+3^2+3^3+...+3^{2020}\)
\(3S-S=\left(3+3^2+3^3+...+3^{2020}\right)-\left(1+3+3^2+...+3^{2019}\right)\)
\(2S=3^{2020}-1\)
\(S=\frac{3^{2020}-1}{2}\)
b, \(S=1+3+3^2+3^3+...+3^{2019}\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\)
\(S=4+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)
\(S=4\cdot1+3^2\cdot4+...+3^{2018}\cdot4\)
\(S=4\left(1+3^2+...+3^{2018}\right)⋮4\)
\(7^{1999}=7^{1996}\cdot7^3=\overline{......1}\cdot\overline{......3}=\overline{......3}\)
\(8^{2015}=8^{2012}\cdot8^3=\overline{.....6}\cdot\overline{......2}=\overline{......2}\)
\(9^{3^2}=9^9=9^8\cdot9=\overline{......1}\cdot\overline{......9}=\overline{.....9}\)
\(87^{32}=\overline{......1}\)
\(58^{33}=58^{32}\cdot58=\overline{.....6}\cdot58=\overline{.....8}\)
2^500 = (2^5)^100 = 32^100
5^200 = (5^2)^100 = 25^100
=> 2^500 > 5^200
3^200 = (3^2)^100 = 9^100
2^300 = (2^3)^100 = 8^100
=> 3^200 > 2^300
Ta có : 2500 =(25)100 =32100
5200 =(52)100 =25100
So sánh : 32100< 25100
Nên : 2500<5200
Câu kia cũng tương tự nhé =))
\(2A=2+2^2+2^3+...+2^{2009}\)
\(A=2A-A=2^{2009}-1=B\)
2A=2+22+23+...+22009
A=2A-A=2^{2009}-1=BA=2A−A=22009−1=B
( x + 1) + (x + 2) + (x+3) +…+ (x + 100) = 7450
=> x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450
=> (x + x + x + .... + x) + (1 + 2 + 3 + ... + 100) = 7450
=> 100x + 5050 = 7450
=> 100x = 1400
=> x = 14
\(A=\left|x+9\right|+\left|6-x\right|+450\)
\(\left|x+9\right|\ge-x-9\)
\(\left|6-x\right|\ge x-6\)
\(\Rightarrow A\ge-x-9+x-6+45\)
\(\Rightarrow A\ge30\)
xét A = 30 khi
\(\hept{\begin{cases}x+9< 0\\6-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -9\\x>6\end{cases}\Rightarrow}voli}\)