\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\) 

Ta thấy rằng :  \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)

Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)

\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)

Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)

\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)

Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)

\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)

Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)

\(\Rightarrow\dfrac{2135}{13790}< 1\)

Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)

\(\Rightarrow\dfrac{4}{3}>1\)

Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)

\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\) 

Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)

\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)

Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)

\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)

\(e)\dfrac{35}{36}và\dfrac{16}{17}\)

Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)

            \(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)

Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)

\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)

\(f)-1,3< -1,2\)

a) Ta có: 

\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)

\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)

Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)

Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)

b) Ta có: 

\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)

\(\dfrac{-60}{-72}=\dfrac{5}{6}\)

Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)

c) Ta có:

\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu) 

\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu) 

Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)

d) Ta có:

\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)

\(\dfrac{10}{9}=\dfrac{1}{9}+1\)

Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)

Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)

e) Ta có:

\(\dfrac{35}{36}=1-\dfrac{1}{36}\)

\(\dfrac{16}{17}=1-\dfrac{1}{17}\)

Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)

Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)

f) Ta có: \(1,3>1,2\)

\(\Rightarrow-1,3< -1,2\)

a)=

b)>

c)=

`17/20 = 119/140`

`11/14=110/140`

`=> 119/140 > 110/140`

`=> 17/20 > 11/14`

\(\dfrac{17}{20}=\dfrac{17\cdot14}{20\cdot14}=\dfrac{238}{280}\)

\(\dfrac{11}{14}=\dfrac{11\cdot20}{14\cdot20}=\dfrac{220}{280}\)

mà 238>220

nên \(\dfrac{17}{20}>\dfrac{11}{14}\)

Ta có:11/12 và 17/18

⇒11x18<12x17 (198<204)

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a) \(A=2A-A\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1-\dfrac{1}{2^{2022}}\)

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )