Question

[Dinh Thi Hai Ha]

1, Cho biểu thức P=\(\dfrac{1000}{100-x}\)với x thuộc Z và x khác 0. Tìm giá trị của x để P có giá trị lớn nhất. T2im giá trị lớn nhất đó của P.

2, Cho các biểu thức:

A=\(\left(\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\right)\)

B=\(\left(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\right)\)

Chứng minh rằng \(\dfrac{A}{B}< \dfrac{1}{8}\)

Answer 1

[Mashiro Shiina]

\(P=\dfrac{1000}{100-x}\)

\(P_{MAX}\Rightarrow P\in Z^+\)

\(\Rightarrow100-x=1\)

\(\Rightarrow x=100-1=99\)

\(\Rightarrow P_{MAX}=\dfrac{1000}{100-99}=1000\)

\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+.....+\dfrac{1}{50.56}\)

\(A=\dfrac{1}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+.....+\dfrac{1}{50}-\dfrac{1}{56}\right)\)

\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)=\dfrac{1}{6}.\dfrac{3}{28}=\dfrac{1}{56}\)

\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)

\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-5\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)

\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}+\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{1}{12}-\dfrac{1}{64}\right)=5.\dfrac{13}{192}=\dfrac{65}{192}\)

\(\dfrac{A}{B}=\dfrac{1}{\dfrac{56}{\dfrac{65}{192}}}=\dfrac{24}{455}\)

\(\dfrac{1}{8}=\dfrac{3}{24}\)

\(\Rightarrow\dfrac{A}{B}< \dfrac{1}{8}\rightarrowđpcm\)