Question

chứng tỏ rằng :\(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)

Answer 1

Đặt vế trái của Bất đẳng thức la A

\(A< \frac{1}{8}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}.\)

\(A< \frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{3}{10}< \frac{5}{10}=\frac{1}{2}\)

Answer 2

hhhhhhhhh

Answer 3

Ta thấy: \(\frac{1}{8}< \frac{1}{2}\)

             \(\frac{1}{11}< \frac{1}{2}\)

              \(\frac{1}{12}< \frac{1}{2}\) 

              \(\frac{1}{13}< \frac{1}{2}\)

              \(\frac{1}{41}< \frac{1}{2}\)

               \(\frac{1}{42}< \frac{1}{2}\)

               \(\frac{1}{43}< \frac{1}{2}\)

=> \(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)