1 + \(\dfrac{19}{120}\)<\(\dfrac{x}{36}\)<\(\dfrac{58}{90}\)+\(\dfrac{59}{72}\)
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ND
Nguyễn Đức Trí
VIP
24 tháng 8 2023
a) \(\dfrac{27}{35}>\dfrac{19}{35}>\dfrac{19}{41}\)
\(\Rightarrow\dfrac{27}{35}>\dfrac{19}{41}\)
b) \(\dfrac{120}{121}< \dfrac{120+1}{121+1}=\dfrac{121}{122}\)
\(\Rightarrow\dfrac{120}{121}< \dfrac{121}{122}\)
DM
16 tháng 3 2017
Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)
Quy đồng mẫu số :
\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)
\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)
Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)
10 tháng 5 2021
`S=1/19+1/19^2+1/19^3+........+1/19^20`
`=>19S=1+1/19+1/19^2+.....+1/19^19`
`=>19S-S=18S=1-1/19^20<1`
`=>S<1/18(đpcm)`
10 tháng 5 2021
Giải:
S=\(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\)
19S=\(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\)
19S-S=\(\left(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\right)-\left(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\right)\)
18S=1-\(\dfrac{1}{19^{10}}\)
S=(1-\(\dfrac{1}{19^{10}}\) ):18
S=\(1:18-\dfrac{1}{19^{10}}:18\)
S=\(\dfrac{1}{18}-\dfrac{1}{19^{10}.18}\)
⇒S<\(\dfrac{1}{18}\) (đpcm)
Chúc bạn học tốt!
LT
Lương Thị Vân Anh
CTVHS
30 tháng 1 2023
\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)
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
HN
3 tháng 5 2017
Ta có: \(\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}=\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+1\)
\(=\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}=20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)\)
Thế lại bài toán ta được
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=\dfrac{20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)
MV
3 tháng 5 2017
Ta có
\(\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}\\ =\dfrac{1}{19}+1+\dfrac{2}{18}+1+\dfrac{3}{17}+1+...+\dfrac{19}{1}+1-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{1}-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+20-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{2}+1+19-19\\ =\dfrac{20}{20}+\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}\\ =20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)\)
Thế vào ta có:
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\\ =\dfrac{20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)}{\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}}\\ =20\)
TT
2
MN
8 tháng 6 2021
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
10 tháng 6 2021
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
=>360+57<10x<58x4+59x5
=>417<10x<527
\(\Leftrightarrow10x\in\left\{420;430;440;...;510;520\right\}\)
hay \(x\in\left\{42;43;44;...;51;52\right\}\)