So sánh: A=\(\dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}\)với 0,9
So sánh S=\(\dfrac{1}{2}\)+\(\dfrac{1}{11}\)+\(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}\)với \(\dfrac{9}{10}\)
Ta có: \(S< \dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{32}\) \(=\dfrac{1}{2}+\dfrac{3}{11}+\dfrac{2}{31}+\dfrac{2}{32}\)
\(=\dfrac{4909}{5456}< \dfrac{9}{10}\)
\(\Rightarrow S< \dfrac{9}{10}\)
Vậy \(S< \dfrac{9}{10}\)
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.
CHỨNG TỎ RẰNG :
\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+\(\dfrac{1}{35}\)+.....+\(\dfrac{1}{148}\)+\(\dfrac{1}{149}\)\(\dfrac{1}{150}\)<\(\dfrac{13}{6}\)
Điền dấu < ; > ; =
3\(\dfrac{1}{5}-\dfrac{8}{3}...........\dfrac{17}{5}-2\dfrac{12}{5}\)
\(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{89}+\dfrac{1}{90}\) ..... \(\dfrac{5}{6}\)
Bài 1:So Sánh(nhanh nhất)
\(\dfrac{-13}{38}va\dfrac{29}{-88},\dfrac{-18}{31}va\dfrac{-1818}{3131}\)
Bài 2:Tính
a)\(\dfrac{-1}{39}+\dfrac{-1}{52}\)
b)\(\dfrac{-6}{9}+\dfrac{-12}{16}\)
Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
So sánh \(\dfrac{1}{11}\)+ \(\dfrac{1}{12}\) +...+ \(\dfrac{1}{19}\)+ \(\dfrac{1}{20}\) với \(\dfrac{2}{3}\)
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 :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)