\(\left(1\right)\) The first thing I did to solve this problem was write out every step, so it  would look something this:

\(6+5.4+4.4+3.4+2.4+1.4\)

\(=6+15.4=6+60=66\)

\(\left(2\right)\)  I was looking at the patterns that I had already found and then realized I needed to find a pattern between \(n\) (the number of cubes high) and the total number of cubes in the tower so I made this chart:

I found that the number being multiplied to get the total number of cubes was going up by odd numbers.  I then noticed that it would be equal to \(2n-1\).  I then found the equation for all n values to be:

Total number of cubes \(=n\left(2n-1\right)\)

Total number of cubes \(=2n^2-n\)

So total number of cubes \(=2n^2-n\)

We have:

\(\left(a\right)2^{600}=\left(2^6\right)^{100}=64^{100}\)

\(\left(b\right)3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(\left(c\right)4^{400}=\left(4^4\right)^{100}=256^{100}\)

\(\left(d\right)5^{300}=\left(5^3\right)^{100}=125^{100}\)

\(\left(e\right)6^{200}=\left(6^2\right)^{100}=36^{100}\)

Because \(36< 64< 125< 243< 256\)

\(\Rightarrow36^{100}< 64^{100}< 125^{100}< 243^{100}< 256^{100}\)

So number  \(\left(e\right)\)  is smallest

The width of the rectangle is:

\(b=\dfrac{S}{a}=\dfrac{30}{5}=6\left(cm\right)\)

The perimeter of the rectangle is:

\(\left(a+b\right)\times2=\left(5+6\right)\times2=11\left(cm\right)\)

Put \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)

We see:

\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow A< 1-\dfrac{1}{n}< 1\)

Conclude:

  \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (The thing must prove)

\(2\left(x^2+\dfrac{1}{x^2}\right)+3\left(x+\dfrac{1}{x}\right)-16=0\left(1\right)\)

Condition: \(x\ne0\)

Put \(t=x+\dfrac{1}{x}\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

\(\left(1\right)\Leftrightarrow2t^2+3t-20=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=\dfrac{5}{2}\end{matrix}\right.\)

If \(t=-4\Rightarrow x=-2\pm\sqrt{3}\)

If \(t=\dfrac{5}{2}\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Conclude:...

Put the equation above is \(\left(1\right)\)

If \(xy>0\) then:

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{y}+\dfrac{2}{x}=2011\\\dfrac{1}{y}-\dfrac{2}{x}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1007}{9}\\\dfrac{1}{x}=\dfrac{490}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{490}\\y=\dfrac{9}{1007}\end{matrix}\right.\) (satisfy)

If  \(xy< 0\) then:

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{y}+\dfrac{2}{x}=-2011\\\dfrac{1}{y}-\dfrac{2}{x}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-\dfrac{1004}{9}\\\dfrac{1}{x}=-\dfrac{1031}{18}\end{matrix}\right.\)\(\Rightarrow xy>0\)  (unsatisfactory)

If \(xy=0\) then: \(\left(1\right)\Leftrightarrow x=y=0\) (satisfy)

Conclude: equations have 2 solutions: \(\left(0;0\right)\) and \(\left(\dfrac{9}{490};\dfrac{9}{1007}\right)\)

\(x\div3\div2=9\)

\(\Rightarrow x=9.3.2\)

\(\Rightarrow x=54\)

Vậy \(x=54\) thì thỏa mãn