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.

Ta có:

\(A=\dfrac{17^{11}-1}{17^{12}-1}< 1\) ( vì tử < mẫu )

\(\Rightarrow A=\dfrac{17^{11}-1}{17^{12}-1}< \dfrac{17^{11}-1+16}{17^{12}-1+16}=\dfrac{17^{11}-17}{17^{12}-17}=\dfrac{17.\left(17^{10}-1\right)}{17.\left(17^{11}-1\right)}=\dfrac{17^{10}-1}{17^{11}-1}=B\)

\(\Rightarrow A< B\)

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

Ta thấy:  

\(\frac{9}{10}< 1< \frac{12}{11}\)

\(\Rightarrow\frac{9}{10}< \frac{12}{11}\)

a / 12 / 11

b/   16/17

-11/12 > 17/-18

\(-\dfrac{11}{12}=\dfrac{-11.3}{12.3}=\dfrac{-33}{36}\)

\(\dfrac{17}{-18}=\dfrac{-17}{18}=\dfrac{-17.2}{18.2}=\dfrac{-34}{36}\)

Do \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)

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

a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)

\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)

vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)

Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)

Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)

Các bài sau tương tự

a: =-5/11-6/11+1=-11/11+1=0

b: =-13/17-13/21-4/17=-1-13/21=-34/21

b: \(=-\dfrac{5}{12}\cdot\dfrac{9}{20}\cdot\dfrac{7}{17}=\dfrac{-21}{272}\)

d: \(=\dfrac{13}{17}\left(-\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{13}{17}\cdot\dfrac{-31}{20}=\dfrac{-403}{340}\)