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\(18:\dfrac{1}{3}=6\)

18x1/3=6

1/3 của 18 = 18/3=6

10+3*3=19

2+7*2=16

               \(f\left(a\right)=\sqrt{25a^2-30a+9}\)

               \(g\left(a\right)=a\)

We have the following:

        \(f\left(a\right)=g\left(a\right)+7\)

\(\Leftrightarrow\sqrt{25a^2-30a+9}=a+7\)

\(\Leftrightarrow\left\{{}\begin{matrix}25a^2-30a+9=\left(a+7\right)^2\\a+7\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\24a^2-44a-40=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\)

Finally, there are two values of a.

\(\dfrac{n-17}{n+23}=\dfrac{n+23-40}{n+23}=1-\dfrac{40}{n+23}\)

\(\Rightarrow\left\{{}\begin{matrix}40⋮\left(n+23\right)\\n+23\ge23\left(n\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(n+23\right)\in\left\{1;2;4;5;8;10;20;40\right\}\\n+23\ge23\end{matrix}\right.\)

\(\Leftrightarrow n+23=40\Leftrightarrow n=17\)

We have: \(f\left(2\right)=5.2+1=11\)

               \(f\left(-1\right)=5.\left(-1\right)+1=-4\)

then: \(a=f\left(2\right)-f\left(-1\right)=11-\left(-4\right)=15\)

Hence: \(g\left(1\right)=a.1+3=15.1+3=18\)

4:2=2cm

4:\(\dfrac{1}{2}\)= 2

So as to have the function be increasing, 2016-m² must be positive. Therefore, 2016 must be greater than m². Since 2016 is a positive number, m has to range from \(-\sqrt{2016}\) to \(\sqrt{2016}\). In other words, m has to be between -44.899 and 44.899. Since m is a whole number, the minimal value of m is -44 and the maximum is 44. To calculate the number of values, we use this formula:
\(N=\frac{44-(-44)}{1}+1=89\)

To conclude, there are 89 values of m such that the function is increasing.

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