a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$

Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$

b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$

Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$

c)

$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ;  $\frac{4}{8} = \frac{1}{2}$

Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$

\(\)

giup tôi với mọi người oi

thứ 4 tôi phải nộp rồi

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đúng 100%

Bài 1:

Ta thấy A < 1

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

Vậy A < B

Bài 2:

Ta thấy C < 1

=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)

Vậy C < D

a 19/18>2005/2004

b 72/73<98/99

a19/18>2005/2004

b72/73<98/99

a) Quy đồng : \(\frac{19}{18}=1\frac{1}{8}\)       \(\frac{2005}{2004}=1\frac{1}{2004}\)

Ta thấy phần nguyên bằng nhau, ta quy đồng phân phân số : 

\(\frac{1}{8}=\frac{501}{4008}\)             \(\frac{1}{2004}=\frac{2}{4008}\)

Vì : 501 > 2 nên \(\frac{501}{4008}>\frac{2}{4008}\) hay \(\frac{19}{18}>\frac{2005}{2004}\)

b. Quy đồng : \(\frac{72}{73}=\frac{7128}{7277}\)             \(\frac{98}{99}=\frac{7154}{7227}\)

Ví : 7128 < 7154 nên \(\frac{7128}{7227}< \frac{7154}{7227}\) hay \(\frac{72}{73}< \frac{98}{99}\)

13/21<9/11

6/17<9/25

7/12>11/-18

tk mk nha mk đang âm điểm

chúc các bn hok tốt ^-^

thank you very much

Ta co  : 72/73 = 1 - 1/73 ; 98/99 = 1 - 1/99

Vì 1/73 > 1/99 suy ra 1 - 1/73 < 1 - 1/99 hay 72/73 < 98/99

Ta có: \(\frac{72}{73}=1-\frac{1}{73}\);\(\frac{98}{99}=1-\frac{1}{99}\)

Vì \(\frac{1}{73}>\frac{1}{99}\Rightarrow1-\frac{1}{73}< 1-\frac{1}{99}\)

                          \(\Rightarrow\frac{72}{73}< \frac{98}{99}\)

Vậy \(\frac{72}{73}< \frac{98}{99}\)

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

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

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

=> A < B

(98^99-1)/(98^98-1)