Bài 1 :Tìm các số nguyên dương n sao cho \(\frac{n^2}{60-n}\) là một số nguyên.
Bài 2: Tím các cặp số nguyên (x,y) sao cho \(\frac{x}{16}-\frac{1}{y}=\frac{1}{32}\)
Bài 3: Cho A=\(\frac{1}{1.21}+\frac{1}{2.22}+\frac{1}{3.23}+...+\frac{1}{80.100}\)
B=\(\frac{1}{1.81}+\frac{1}{2.82}\frac{1}{3.83}+...+\frac{1}{20.100}\) Tính:\(\frac{A}{B}\)
Bài 4:Tính giá trị biểu thức:
A=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)
3. + \(20A=\frac{21-1}{1\cdot21}+\frac{22-2}{2\cdot22}+...+\frac{100-80}{80\cdot100}\)
\(\Rightarrow20A=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+...+\frac{1}{80}-\frac{1}{100}\)
\(\Rightarrow20A=\left(1+\frac{1}{2}+...+\frac{1}{80}\right)-\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{1}{20}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
+ \(80B=\frac{81-1}{1\cdot81}+\frac{82-2}{2\cdot82}+...+\frac{100-2}{20\cdot100}\)
\(=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}+...+\frac{1}{20}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
\(\Rightarrow B=\frac{1}{80}\left[\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\right]\)
Do đó : \(\frac{A}{B}=\frac{\frac{1}{20}}{\frac{1}{80}}=4\)
4. + \(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}=\frac{100}{2}=50\)