The speed of Jack if the train he is on travels 20 miles in just 6 minutes is:

20 : 6 = 3,3333... ( m/minute )

So the speed of Jack if the train he is on travels 20 miles in just 6 minutes is 3,3333... m/minute

We have: \(\dfrac{x^2\left(x-3\right)}{x-9}< 0\)

\(\Rightarrow x^2\left(x-3\right)< 0\)

But \(x^2\ge0\) so \(x-3< 0\)

\(\Rightarrow x< 3\)

So \(x< 3\)

\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+2\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{\left(n-1\right)n\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)}\right)\)

\(=\dfrac{1}{4}.\dfrac{1}{2n\left(n+2\right)}\)

\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n\left(n+2\right)}< \dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

So \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\)

\(\left(x+2\right)\left(x+2\right)=24\)

\(\Rightarrow x^2+4x+4=24\)

\(\Rightarrow x^2+4x=20\)

We have the following table:

...

We have: \(\left|a\right|=2\Rightarrow a=2\) or a = -2

\(\left|b\right|=3\Rightarrow b=3\) or b = -3

\(\left|c\right|=4\Rightarrow c=4\) or c = -4

\(\Rightarrow\left(a;b;c\right)\in\left(2;3;4\right);\left(-2;-3;-4\right)\)

but \(a>b>c\)

\(\Rightarrow\left(a;b;c\right)\in\left(-2;-3;-4\right)\)

\(\Rightarrow a+b-c=-2-3+4=-1\)

So the value of a + b - c = -1

Change \(x=\dfrac{1}{2}\) into \(4x-3a=0\) , we have:

\(2-3a=0\)

\(\Rightarrow3a=2\)

\(\Rightarrow a=\dfrac{2}{3}\)

So \(a=\dfrac{2}{3}\)

Put \(A=\left|m-2\right|+\left|m+3\right|\)

We have: \(A=\left|m-2\right|+\left|m+3\right|=\left|2-m\right|+\left|m+3\right|\)

Apply the inequality \(\left|a\right|+\left|b\right|\ge\left|a\right|+\left|b\right|\), we have:

\(A\ge\left|2-m+m+3\right|=\left|5\right|=5\)

The "=" sign occurs when \(\left\{{}\begin{matrix}2-m\ge0\\m+3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m\le2\\m\ge-3\end{matrix}\right.\Rightarrow-3\le m\le2\)

So \(MIN_A=5\) when \(-3\le m\le2\)