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\(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\\ =\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+...+\dfrac{1}{\dfrac{59\cdot60}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{59\cdot60}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =2\cdot\dfrac{19}{60}\\ =\dfrac{38}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm
\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{13}{3}.\dfrac{-1}{3}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)
\(\dfrac{-13}{9}\)\(\le x\le\)\(\dfrac{-11}{18}\)
\(\dfrac{-26}{18}\)\(\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x\in\left\{\dfrac{-12}{18};\dfrac{-13}{18};\dfrac{-14}{18};\dfrac{-15}{18};...;\dfrac{-24}{18};\dfrac{-25}{18}\right\}\)Tick hộ mình nha bạn
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}< x< \dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}< x< \dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}\)
\(\Leftrightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(\Leftrightarrow-2< 24x< 3\)
=>x=0
b: \(\Leftrightarrow\dfrac{9-10}{12}< \dfrac{x}{12}< 1-\dfrac{8-3}{12}=\dfrac{7}{12}\)
=>-1<x<7
hay \(x\in\left\{0;1;2;3;4;5;6\right\}\)
Ta có \(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\)
= \(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\dfrac{1}{3}-\dfrac{1}{60}=\dfrac{19}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
Vậy M < \(\dfrac{2}{3}\)
Ta có: