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Ta có:
\(B=2^{2012}+2^{2011}+...+2^3+2^2+2+1\)
\(\Rightarrow2B=2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\)
\(\Rightarrow2B-B=\left(2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\right)-\left(2^{2012}+...+1\right)\)
\(\Rightarrow B=2^{2013}-1\)
\(A=2^{2003}.9+2^{2003}.1005\)
\(\Rightarrow A=2^{2003}.\left(9+1005\right)\)
\(\Rightarrow A=2^{2003}.1024\)
\(\Rightarrow A=2^{2003}.2^{10}\)
\(\Rightarrow A=2^{2013}\)
Vì \(2^{2013}-1< 2^{2013}\) nên A > B
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=3^2+6^2+9^2+...+30^2\)
\(=\left(1.3\right)^2+\left(2.3\right)^2+\left(3.3\right)^2+...+\left(10.3\right)^2\)
\(=\left(1^2+2^2+3^2+...+10^2\right).3^2\)
\(=385.9\)
\(=3465\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
3^2 = 1^2.3^2 = 1^2.9
6^2 = 2^2.3^2 = 2^2.9
...
42^2 = 14^2.3^2 = 14^2.9
--> 3^2 + 6^2 + ... + 42^2 = (1^2 + 2^2 + ... + 14^2).9 = 1015.9 = 9135
1015.9 = 9135