Có A = 17^11 - 1/17^12 - 1

         = 17^10 . 17 - 1/17^11 . 17 - 1

Có B = 17^10 + 1/17^11 + 1

 Vì 17^11 > 17^10 

Lại có : 1/17^12 < 1/17^11

Mà: 17 > 1/17

=> A > B

A>B                        NHỚ KICK NHA BẠN 

MK cần cả cách trình bày cơ bạn

chụi

ta có :

A=17^11-1/17^12-1

  =17^10.17-1/17^11.17-1

B=17^10+1/17^11+1

ta thấy 17^11>17^10

mà 1/17^12<1/17^11

lại có 17>1/17

nên=>A>B

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

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) Ta có :

\(\frac{7}{9}< 1\); \(\frac{19}{17}>1\)

Vì \(\frac{7}{9}< 1< \frac{19}{17}\)nên \(\frac{7}{9}< \frac{19}{17}\)

b) Xét phân số trung gian là \(\frac{n}{n+2}\)

Vì \(\frac{n}{n+3}< \frac{n}{n+2}\)và \(\frac{n}{n+2}< \frac{n+1}{n+2}\)

\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)

c) Ta có :

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)

Vậy \(A< B\)

a)7/9<1,19/17 => 7/9<19/17.

\(A=\frac{17^{18}+1}{17^{19}+1}\)

\(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{\left(17^{18}+1\right)+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(\text{Vì}\)\(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

Trả lời

\(17A=\frac{\left(17^{18}+1\right)17}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(17B=\frac{\left(17^{17}+1\right)17}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Vì \(17^{19}+1>17^{18}+1\)

\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)

\(\Rightarrow B>A\)