Ta có : \(\dfrac{2n+5}{n-3}=\dfrac{2n-6+11}{n-3}\)

\(=\dfrac{2n-6}{n-3}+\dfrac{11}{n-3}\)

\(=\dfrac{2\left(n-3\right)}{n-3}+\dfrac{11}{n-3}=2+\dfrac{11}{n-3}\)

Để giá trị trên là số nguyên \(\Leftrightarrow\dfrac{11}{n-3}\in Z\)

\(\Leftrightarrow\left(n-3\right)\inƯ\left(11\right)\)

mà \(Ư\left(11\right)=\left(\pm1;\pm11\right)\)

\(\Rightarrow n\in\left(4;2;14;-9\right)\)

\(Vậy....\)

3

vì số đầu là tổng của 2 số sau

\(\dfrac{-2}{6}\)

\(\Leftrightarrow\left(x+3\right)^7\left[\left(x+3\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x+3\right)^7\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-4\end{matrix}\right.\)

Vậy...

`(4x+1)(-x-9)=0`

\(< =>\left[{}\begin{matrix}4x+1=0\\-x-9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}4x=-1\\-x=9\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-9\end{matrix}\right.\)

`(4x+1)(-x-9)=0`

TH1: `4x+1=0`

`=>4x=-1`

`=>x=-1/4`

TH2: `-x-9=0`

`=>-x=9`

`=>x=-9`

Vậy `x\in{-1/4;-9}`

Lời giải:

$=\frac{81}{20}.\frac{4}{9}+\frac{4}{15}: \frac{-4}{7}$

$=\frac{9.9.4}{4.5.9}+\frac{4}{15}.\frac{-7}{4}$

$=\frac{9}{5}+\frac{-7}{15}=\frac{27}{15}+\frac{-7}{15}=\frac{20}{15}=\frac{4}{3}$