Cho :
\(S=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)
Hãy so sánh S và \(\dfrac{1}{2}\)
Cho S = \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\). Hãy so sánh S và \(\dfrac{1}{2}\)
Ta có: \(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
\(\dfrac{1}{13}>\dfrac{1}{20}\)
\(\dfrac{1}{14}>\dfrac{1}{20}\)
\(\dfrac{1}{15}>\dfrac{1}{20}\)
\(\dfrac{1}{16}>\dfrac{1}{20}\)
\(\dfrac{1}{17}>\dfrac{1}{20}\)
\(\dfrac{1}{18}>\dfrac{1}{20}\)
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
=> \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}.10\)
hay S > \(\dfrac{1}{2}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 11 < 20 )
\(\dfrac{1}{12}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 12 < 20 )
...
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)( 10 số hạng )
\(\Rightarrow S>\dfrac{1}{20}.10\Rightarrow S>\dfrac{10}{20}\Rightarrow S>\dfrac{1}{2}\)
Vậy ...
Ta có các phân số trên đều lần lượt lớn hơn \(\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>\dfrac{1}{20}.10\)\(\Rightarrow A>\dfrac{1}{20}.10\)
\(\Rightarrow A>\dfrac{10}{20}\)
\(\Rightarrow A>\dfrac{1}{2}\)
Vậy \(A>\dfrac{1}{2}\)
cho A =\(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)SO SÁNH A VỚI \(\dfrac{1}{2}\)
Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)
=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)
=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)
Vậy...
Cho S =\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)
Hãy so sánh S và \(\dfrac{1}{2}\)
Các bạn giúp mình nhé mình đang cần gấp
ta thấy : \(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...\dfrac{1}{19}\)đều lớn hơn\(\dfrac{1}{20}\)
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)(20 số hạng \(\dfrac{1}{20}\))
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>1\) mà 1 > \(\dfrac{1}{2}\) =>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>\dfrac{1}{2}\)
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{2}{18}+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}}\)
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
Biến đổi tử số
\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)
= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)
= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)
= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)
Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)
Vậy A = 20
So sánh phân số \(\dfrac{2012}{2013}\) với A, biết:
A = \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{13}\) + \(\dfrac{1}{14}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{38}\) + \(\dfrac{1}{39}\) + \(\dfrac{1}{40}\)
Ai xong đầu tiên mình tick nhé.
Cho S = \(\dfrac{6}{15}+\dfrac{6}{16}+\dfrac{6}{17}+\dfrac{6}{18}+\dfrac{6}{19}\) Chứng minh rằng 1<S<2
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
Tính hợp lí :
a) \(\dfrac{14}{13}\) + ( \(\dfrac{-1}{13}\) - \(\dfrac{19}{20}\) )
b) \(\dfrac{-24}{17}\) - ( \(\dfrac{-7}{17}\) - \(\dfrac{1}{16}\) )
a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)
b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)
\(\dfrac{14}{13}+\left(\dfrac{-1}{13}-\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=\\ \dfrac{13}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\)
Câu b em làm tương tự nhé
\(a.\dfrac{14}{13}+\left(\dfrac{-1}{13}+\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=1+\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\\ b.\dfrac{-24}{17}+\left(\dfrac{-7}{17}-\dfrac{1}{16}\right)=\left(\dfrac{-24}{17}+\dfrac{7}{17}\right)+\dfrac{1}{16}=-1+\dfrac{1}{16}=\dfrac{-16}{16}+\dfrac{1}{16}=\dfrac{-15}{16}\)
Bài 1: Thực hiện phép tính:
1) \(\dfrac{-17}{30}-\dfrac{11}{-15}+\dfrac{-7}{12}\)
2) \(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
3) \(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}-\dfrac{18}{25}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
So sánh \(\dfrac{1}{11}\)+ \(\dfrac{1}{12}\) +...+ \(\dfrac{1}{19}\)+ \(\dfrac{1}{20}\) với \(\dfrac{2}{3}\)