Ta có : y = \(\dfrac{5x+9}{x+3}\)

Để y nhận giá trị nguyên thì: 5x + 9 \(⋮\) x + 3

=> 5. ( x + 3 ) + 9 - 15 \(⋮\) x + 3

=> 5. ( x + 3 ) - 6 \(⋮\) x + 3

=> 6 \(⋮\) x + 3 ( vì 5. ( x + 3 ) \(⋮\) x + 3 )

=> x + 3 \(\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

=> \(x\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\)

Vậy : \(x\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\) thì y nhận giá trị nguyên.

\(y=\frac{5x+9}{x+3}=\frac{5x+15-6}{x+3}=\frac{5\left(x+3\right)-6}{x+3}=5-\frac{6}{x+3}\)

y nguyên khi \(\frac{6}{x+3}\) nguyên <=> 6 chia hết cho x+3

<=>\(x+3\inƯ\left(6\right)=\){-6;-3;-2;-1;1;2;3;6}

<=> \(x\in\){-9;-6;-5;-4;-2;-1;0;3}

 \(A=2018-\left|x-7\right|-\left|y+2\right|\)

Ta có: \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\\left|y+2\right|\ge0\forall y\end{cases}}\Rightarrow2018-\left|x-7\right|-\left|y+2\right|\le2018\)

\(A=2018\Leftrightarrow\hept{\begin{cases}\left|x-7\right|=0\\\left|y+2\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}}\)

Vậy \(A_{m\text{ax}}=2018\Leftrightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)

Tham khảo~

Đáp án là 

99 x 100 x 101 = 999900

Vậy A = 999900

Bạn ghi bị lộn đề rồi, hai số cuối phải là \(97.98.99\)và \(98.99.100\)

\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+97.98.99.4+98.99.100.4\)

\(=1.2.3.4+2.3.4\left(5-1\right)+3.4.5\left(6-2\right)+...+97.98.99\left(100-96\right)+98.99.100.\left(101-97\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100\)

\(=98.99.100.101=97990200\Rightarrow A=\frac{97990200}{4}=24497550\)

A = 2004.2005
= (2000 + 4).2005
= 2000.2005 + 4.2005
B = 2000.2009
= 2000.(2005 + 4)
= 2000.2005 + 2000.4
=> A > B
@Nguyễn Hoàng Nina