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)\(12< 13;49>47\)

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

b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)

c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)

d)

\(1-\dfrac{67}{77}=\dfrac{10}{77}\)

\(1-\dfrac{73}{83}=\dfrac{10}{83}\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)

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

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

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

b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)

\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)

c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)

d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)

\(\dfrac{73}{83}+\dfrac{10}{83}=1\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)

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

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)

các bạn giúp mk vs

mk đg cần gấp

Ta có :

\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)

a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)

\(\dfrac{-5}{6}>\dfrac{-91}{104}\)

Sao bn giống BT mình thế ?:)

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)