Question

 chứng tỏ rằng  A >1 với

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

      AI NHANH MK TICK CHO NHOA ! 

(MK CÁM ƠN)

 

Answer 1

Ta có : 

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

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

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

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

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

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

\(\Rightarrow\)\(A>1\)

Vậy \(A>1\)

Chúc bạn học tốt ~ 

Answer 2

Ko cần dài dòng vậy đâu,A=\(\frac{5^2}{1.6}+\left(\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\right)\)

Ta thấy \(\frac{5^2}{1.6}>1\)và tổng trong ngoặc >0  nên =>A>1

Answer 3

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

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

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

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

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

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

\(\text{Vì }\frac{150}{31}>1\Rightarrow A>1\text{(đpcm)}\)

Chúc bạn học tốt !!!

Answer 4

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

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

\(\Leftrightarrow A=5\left(1-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}.\)