a: \(\dfrac{12}{49}< \dfrac{13}{49}< \dfrac{13}{47}\)

b: \(\dfrac{12}{47}>\dfrac{19}{47}>\dfrac{19}{77}\)

a/ so sánh bằng 12/47

b/ chưa tìm ra

a) Ta chọn phân số \(\dfrac{12}{47}\) là phân số trung gian, ta có :

\(\dfrac{12}{49}< \dfrac{12}{47};\dfrac{13}{47}>\dfrac{12}{47}\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)

p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)

\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)

\(x=\dfrac{2}{7}\)

2/7 nha

hok tốt

k mik đi

Ta có : \(\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4.\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)

\(=\dfrac{3\times\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\times\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}\)

\(=\dfrac{3}{4}\)

\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

Mình chỉ cho đáp án thôi,sai thì châm chước cho mìn nha!

a)-91 phần200

b)-25phần 4

c)5 phần 2

d)2

e)0

a, ( 0,36-2,18) : ( 3,8 + 0,2)

= -1,82 : 4

=-0,455 hay -91/200

b, 3/8*19/1/3-3/8*33/1/3

=3/8*(19/1/3-33/1/3)

=3/8*(-14)

=-21/4

sorry

tính nhanh

a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

a) gọi biểu thức đó là A

Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Dựa vào công thức trên, ta có

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)

\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)

b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)

a) \(\dfrac{12}{47}\) và \(\dfrac{11}{53}\)

Ta có: \(\dfrac{11}{47}>\dfrac{11}{53}\) mà \(\dfrac{12}{47}>\dfrac{11}{47}\)

\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)

a) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{11}{44}>\dfrac{11}{53}\)

\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)

b) Ta có : \(\dfrac{456}{461}=1-\dfrac{5}{461}\)

\(\dfrac{123}{128}=1-\dfrac{5}{128}\)

Vì \(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow1-\dfrac{5}{461}>1-\dfrac{5}{128}\)

\(\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

c) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{19}{76}>\dfrac{19}{77}\)

=> \(\dfrac{12}{47}>\dfrac{19}{77}\)

d) Ta có : \(13A=13.\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{16}+13}{13^{16}+1}=\dfrac{13^{16}+1+12}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)

\(13B=13.\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{17}+13}{13^{17}+1}=\dfrac{13^{17}+1+12}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)

Ta thấy : \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\Rightarrow\dfrac{13^{15}+1}{13^{16}+1}>\dfrac{13^{16}+1}{13^{17}+1}\)

a) Ta có : \(\dfrac{12}{47}>\dfrac{12}{53}>\dfrac{11}{53}\) \(\Leftrightarrow\dfrac{12}{47}>\dfrac{11}{53}\) b) Ta có : \(\dfrac{456}{461}=\dfrac{461-5}{461}=1-\dfrac{5}{461}\) \(\dfrac{123}{128}=\dfrac{128-5}{128}=1-\dfrac{5}{128}\) Do \(1-\dfrac{5}{461}>1-\dfrac{5}{128}\) \(\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\) c) Ta có: \(\dfrac{12}{47}\) > \(\dfrac{12}{48}=\dfrac{1}{4}\) \(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\) Do \(\dfrac{12}{47}>\dfrac{1}{4}>\dfrac{19}{77}\) \(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\) d) Ta có : A=\(\dfrac{13^{15}+1}{13^{16}+1}\) \(\Leftrightarrow\) 13A=\(\dfrac{13.\left(13^{15}+1\right)}{13^{16}+1}\) \(\Leftrightarrow\) 13A=\(\dfrac{13^{16}+13}{13^{16}+1}\) \(=\dfrac{13^{16}+1+12}{13^{16}+1}=\dfrac{13^{16}+1}{13^{16}+1}+\dfrac{12}{13^{16}+1}\)
\(=1+\dfrac{12}{13^{16}+1}\) B=\(\dfrac{13^{16}+1}{13^{17}+1}\) \(\Leftrightarrow\) 13B=\(\dfrac{13.\left(13^{16}+1\right)}{13^{17}+1}\)

\(\Leftrightarrow\) 13B=\(\dfrac{13^{17}+13}{13^{17}+1}=\dfrac{13^{17}+1+12}{13^{17}+1}\) \(=\dfrac{13^{17}+1}{13^{17}+1}+\dfrac{12}{13^{17}+1}\) \(=1+\dfrac{12}{13^{17}+1}\)

Do \(1+\dfrac{12}{13^{16}+1}.>1+\dfrac{12}{13^{17}+1}\) nên 13A>13B \(\Rightarrow\) A>B