Chứng tỏ rằng :\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{2}{3}\)
Tính tích \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{899}{900}\)
Chứng tỏ rằng : \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
Tính giá trị của biểu thức sau :
\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...
A=3/22.8/32.15/42......899/302 A=3.8.15.....899/22.32.42.....302 A=(1.3).(2.4).(3.5).....(29.31)/(2.3.4....30)(2.3.4...30) A=(1.2.3....29).(3.4.5...31)/(2.3.4...30)(2.3.4...30) A=1.31/30.2=31/60
+ Theo mk thì: Vì \(\dfrac{1}{5}\) ,\(\dfrac{1}{6}\),\(\dfrac{1}{7}\), ... ,\(\dfrac{1}{17}\) < 1
Mà 1 < 2
=>\(\dfrac{1}{5}\) ,\(\dfrac{1}{6}\),\(\dfrac{1}{7}\), ... ,\(\dfrac{1}{17}\) < 2
+ Mk nhớ ko nhàm, thì:
\(M=(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})+(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5})+...+(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12})\)
xong thì m sẽ chỉ còn 1/1+1/12 thôi
Vậy cậu tính m đc ùi
