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Giải:
a)
\(\dfrac{7}{48}=\dfrac{105}{720};\)
\(\dfrac{11}{72}=\dfrac{110}{720};\)
\(\dfrac{17}{120}=\dfrac{102}{720}\)
Vì \(102< 105< 110\)
\(\Leftrightarrow\dfrac{102}{720}< \dfrac{105}{720}< \dfrac{110}{720}\)
\(\Leftrightarrow\dfrac{17}{120}< \dfrac{7}{48}< \dfrac{11}{72}\)
Vậy ...
b) \(\dfrac{31}{49}=\dfrac{60140}{95060};\)
\(\dfrac{62}{97}=\dfrac{60760}{95060};\)
\(\dfrac{93}{140}=\dfrac{63147}{95060}\)
Vì \(60140< 60760< 63147\)
\(\Leftrightarrow\dfrac{60140}{95060}< \dfrac{60760}{95060}< \dfrac{63147}{95060}\)
\(\Leftrightarrow\dfrac{31}{49}< \dfrac{62}{97}< \dfrac{93}{140}\)
Vậy ...
a ) \(\dfrac{7}{48}\) = \(\dfrac{105}{720}\)
\(\dfrac{11}{72}\) = \(\dfrac{110}{720}\)
\(\dfrac{17}{120}\) = \(\dfrac{102}{720}\)
Vì 102 < 105 < 110
\(\Leftrightarrow\) \(\dfrac{102}{720}\) < \(\dfrac{105}{720}\) < \(\dfrac{110}{720}\)
\(\Leftrightarrow\) \(\dfrac{17}{120}\) < \(\dfrac{7}{48}\) < \(\dfrac{11}{72}\)
Vậy .....................
( k cho tớ nha . Tớ chỉ bt lm phần a )
50A=\(\left(\frac{49}{1}+.......+\frac{1}{49}\right)49:2\)
50A= 1201
A=1201:50
A=\(\frac{1201}{10}\)=120.1
mà 120,1 ko phải số tự nhiên mà là số thập phân
=>A ko là số tự nhiên
tử số =18.123+9.4567.2+3.5310.6=18.123+18.4567+18.5310
=18(123+4567+5310)
=18.10000=180000
mẫu số = 1+4+7+....+49+52+55+58-490
=(1+58)+(4+55)+.....(28+31)-490
=59+59+59+...+59-490(10 so 59)
=100
suy ra A=180000:100=1800
a/ \(\frac{15}{-50}=-\frac{3}{10}\)
\(\frac{7}{10}\)
\(\frac{24}{-20}=-\frac{12}{10}\)
\(\Rightarrow-\frac{12}{10};-\frac{3}{10};\frac{7}{10}\)
KL: ..........................................
b/ \(\frac{17}{39}=\frac{17.20}{39.20}=\frac{340}{780}\)
\(\frac{11}{65}=\frac{11.12}{65.12}=\frac{132}{780}\)
\(\frac{9}{52}=\frac{9.15}{52.15}=\frac{135}{780}\)
\(\Rightarrow\frac{132}{780};\frac{135}{780};\frac{340}{780}\)
KL:.............................................
c/ \(\frac{17}{20}=\frac{17.9}{20.9}=\frac{153}{180}\)
\(\frac{-19}{30}=-\frac{19.6}{30.6}=-\frac{114}{180}\)
\(\frac{38}{45}=\frac{38.4}{45.4}=\frac{152}{180}\)
\(\frac{-13}{18}=-\frac{13.10}{18.10}=-\frac{130}{180}\)
\(\Rightarrow-\frac{130}{180};-\frac{114}{180};\frac{152}{180};\frac{153}{180}\)
KL:..................................
Sắp xếp theo thứ tự tăng dần đó là \(\frac{31}{49};\frac{62}{97};\frac{93}{140}\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)