Question

a) Tìm số nguyên a sao cho A=\(\dfrac{a^3+3a^2+2a-3}{a+1}\) có giá trị nguyên

b) Cho B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......+\dfrac{1}{9^2}\). Chứng minh rằng: \(\dfrac{8}{9}>B>\dfrac{2}{5}\)

Answer 1

b)B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)

B<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

B<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

B<\(1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)-\dfrac{1}{9}\)

B<1-\(\dfrac{1}{9}\)

B<\(\dfrac{8}{9}\)(1)

ta có:

B>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

B>​\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{10}\)

B>\(\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)...+\left(\dfrac{1}{9}+\dfrac{1}{9}\right)-\dfrac{1}{10}\)

B>\(\dfrac{1}{2}-\dfrac{1}{10}\)

B>\(\dfrac{2}{5}\)