b. S=(2+4+6+...+19+20)^2

P=(3+6+9+...+30)^2

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

\(P=3^2+6^2+9^2+...+30^2\)

     \(=\left(1.3\right)^2+\left(2.3\right)^2+\left(3.3\right)^2+...+\left(10.3\right)^2\)

     \(=\left(1^2+2^2+3^2+...+10^2\right).3^2\)

     \(=385.9\)

     \(=3465\)

|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

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

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