Thao quy ước của 1 phân số lớn hơn 0 thì:
\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\ne0\right)\)
Áp dụng vào từng phân số trên ta có: ( các phân số trên lớn hơn 0 nên):
để ý rằng các phân số trên đều lớn hơn 1/100
=>tích cũng lớn hơn 1/100
=>A>1/100
CHÚC BẠN HỌC TỐT.............
Bài 1:Tính
a, A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot....\cdot\dfrac{9999}{10000}\)
b,B=\(\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot....\cdot\left(1-\dfrac{1}{1326}\right)\)
c,C=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
\(\dfrac{1}{7}\cdot2\dfrac{1}{3}+\dfrac{5}{2}\cdot\dfrac{3}{7}-\dfrac{59}{6}\cdot\dfrac{1}{7}\)
giup mink nha
c) \(\left(6-\dfrac{14}{5}\right)\cdot\dfrac{25}{8}-\dfrac{18}{5}\cdot\dfrac{1}{4}\)
cho :
\(\dfrac{\dfrac{2}{3}n+\dfrac{1}{5}\cdot\dfrac{3}{7}+\dfrac{1}{7}\cdot\dfrac{3}{10}+\dfrac{1}{3}n-\dfrac{1}{14}+\dfrac{33}{35}}{\dfrac{2}{3}\cdot\left(3n+\dfrac{3}{5}\right)\dfrac{14}{15}+\dfrac{1}{3}}\)
a, Hãy rút gọn A.
b,Tìm giá trị của A khi n =\(\dfrac{-1}{5}\)
c,Tìm n để A nhận giá trị là \(\dfrac{2}{5}\)
d,Tìm n để 2A thuộc Z
\(\dfrac{-5}{9}+1\dfrac{5}{9}\cdot\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\)
\(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{104}{195}+25\%\right)\cdot\dfrac{24}{47}-3\dfrac{12}{13}:3\)
\(1\dfrac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
Thực hiện phép tính bằng cách hợp lí :
\(B=1\dfrac{6}{41}\cdot\left(\dfrac{12+\dfrac{12}{9}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\cdot\dfrac{124242423}{237373735}\)
a) \(\dfrac{2}{15}\cdot x-x=30\%\)
b) \(\dfrac{x}{2}\cdot\dfrac{x}{3}=\dfrac{5}{6}\)
tính tổng
S=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\)
A=1+\(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
B= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...\dfrac{1}{100}\)
C=\(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
\(A=\left(1-\dfrac{1}{2}\right)\cdot1-\dfrac{1}{3}\cdot......\cdot\left(1-\dfrac{1}{2016}\right)\cdot\left(1-\dfrac{1}{2017}\right)\)