\(\dfrac{n+5}{n+1}=\dfrac{n+4+1}{n+1}\\=\dfrac{n+1}{n+1}+\dfrac{4}{n+1}\\ =1+\dfrac{4}{n+1}\)

Để `n+5 vdots n+1` thì 

`4 vdots n+1`

`=>n+1 in Ư(5)={+-1;+-5}`

`=> n in {0;-2;4;-6}`

Câu 2:

\(=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{98\cdot100}\right)\\ =\dfrac{1\cdot3+1}{1\cdot3}\cdot\dfrac{2\cdot4+1}{2\cdot4}\cdot\dfrac{3\cdot5+1}{3\cdot5}...\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot\dfrac{16}{3\cdot5}\cdot...\cdot\dfrac{9801}{98\cdot100}\\ =\dfrac{2\cdot3\cdot4\cdot...\cdot99}{1\cdot2\cdot3\cdot...\cdot98}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot99}{3\cdot4\cdot5\cdot...\cdot100}\\ =99\cdot\dfrac{2}{100}=\dfrac{198}{100}\)

\(25\cdot48+25\cdot51+300+25\\ =25\cdot48+25\cdot51+25\cdot12+25\cdot1\\ =25\cdot\left(48+51+12+1\right)\\ =25\cdot112=2800\)

\(\left(318-x\right)+264=327\\ 318-x+264=327\\ \left(318+264\right)-x=327\\ 582-x=327\\ x=582-327\\ x=255\)

\(1-5+9-13+17-...+201-205\\ =\left(1+9+17+201\right)-\left(5+13+...+205\right)\\ =M-N\left(1\right)\)

Xét \(M=1+9+17+...+201\left(26sh\right)\\ =\dfrac{\left(201+1\right)\cdot26}{2}=2626\left(2\right)\)

Xét \(N=5+13+...+205\left(26sh\right)\\ =\dfrac{\left(205+5\right)\cdot26}{2}=2730\left(3\right)\)

Từ (1);(2) và (3)\(\Rightarrow2626-2730=-104\)