a) \(\dfrac{5}{9}< \dfrac{7}{9}\)

b) \(\dfrac{7}{6}>\dfrac{6}{6}\)

c) \(\dfrac{3}{14}< \dfrac{5}{14}\)

d) \(\dfrac{5}{8}< \dfrac{9}{8}\)

a) \(\dfrac{6}{14}=\dfrac{6:2}{14:2}=\dfrac{3}{7}\)

\(\dfrac{3}{7}< \dfrac{4}{7}\)

b) \(\dfrac{6}{15}=\dfrac{6:3}{15:3}=\dfrac{2}{5}\)

\(\dfrac{3}{5}>\dfrac{2}{5}\)

c) \(\dfrac{10}{18}=\dfrac{10:2}{18:2}=\dfrac{5}{9}\)

\(\dfrac{5}{9}>\dfrac{2}{9}\)

125/90=25/18

84/91=12/13

75/45=25/15=5/3

125/90=25/18

84/91=12/13

75/45=25/15=5/3

6/7=18/21=54/63=104/126

125/90 = 25/18 

84/91 = 12/13

75/45 = 25/15 = 5/3

6/7 = 18/21 = 54/63 = 108/126

\(\dfrac{1}{4444}< 1,\dfrac{3}{7}< 1,\dfrac{9}{5}>1,\dfrac{7}{3}>1,\dfrac{14}{15}< 1,\dfrac{16}{16}=1,\dfrac{14}{11}>1\)

1/4 < 1

3/7 < 1

9/5 > 1

7/3 > 1

14/15 < 1

16/16 = 1

14/11 >1

\(A=\dfrac{14^{14}+1}{14^{15}+1}\)

\(\Rightarrow14.A=\dfrac{14^{15}+14}{14^{15}+1}\)

\(\Rightarrow14.A=\dfrac{14^{15}+1}{14^{15}+1}+\dfrac{13}{14^{15}+1}\)

\(\Rightarrow14.A=1+\dfrac{13}{14^{15}+1}\)

\(B=\dfrac{14^{15}+1}{14^{16}+1}\)

\(\Rightarrow14.B=\dfrac{14^{16}+14}{14^{16}+1}\)

\(\Rightarrow14.B=\dfrac{14^{16}+1}{14^{16}+1}+\dfrac{13}{14^{16}+1}\)

\(\Rightarrow14.B=1+\dfrac{13}{14^{16}+1}\)

Nhận xét: \(\dfrac{13}{14^{15}+1}>\dfrac{13}{14^{16}+1}\) (cùng tử, xét mẫu)

\(\Rightarrow A>B\)

Vậy \(A>B\)

a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)

= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)

= 0 + \(\dfrac{11}{125}\)

= \(\dfrac{11}{125}\)

b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +

\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4

= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4

= -1

c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)

= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)

= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)

= \(\dfrac{-7}{75}\)

d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)

= 1 + (-1)

= 0

a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)

Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)

b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)

\(\dfrac{25}{49}=1-\dfrac{24}{49}\)

Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)

Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)

23/27 lớn hơn 22/29

15/25 lớn hơn 25/49

Ta có

Ta có: \(\dfrac{2002}{2003}< \dfrac{14}{13}\) vì \(\dfrac{2002}{2003}< 1\), \(\dfrac{14}{13}>1\)

ta có

2002/2003<1<14/13

=>2002/2003<14/13

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)