Question

A=\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+....+\frac{5^2}{26.31}\)           chứng tỏ A>1    giúp em mk với cảm ơn

 

Answer 1

\(A=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(A=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(A=5.\left(1-\frac{1}{31}\right)\)

\(A=5.\frac{30}{31}\)

\(A=\frac{150}{31}>1\)

Answer 2

Đề hơi lạ nhỉ, vì quá rõ ràng rùi 5²/1.6 = 25/6 > 1 nên A lớn hơn 1

Answer 3

\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

   \(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

   \(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

   \(=5\left(1-\frac{1}{31}\right)\)

    \(=5.\frac{30}{31}=\frac{150}{31}>1\)

=> A > 1

Study well ! >_<

Answer 4

\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5\left(1-\frac{1}{31}\right)\)

\(=5.\frac{30}{31}\)

\(=\frac{150}{31}\)