\(A=\frac{15}{19}\left(\frac{17}{23}+\frac{19}{23}-\frac{13}{23}\right)=\frac{15}{19}\left(\frac{17+19-13}{23}\right)=\frac{15}{19}\cdot\frac{23}{23}=\frac{15}{19}.1=\frac{15}{19}\)

Bài 1

a: 8/15; 11/15; 15/15; 16/15; 17/15; 19/15

b: 19/42; 19/35; 19/21; 19/19; 19/17

c: 8/10<6/7<16/14<37/35

15/19 . 17/19 + 15/19 . 10/9 - (-7/19)

= 15/19 . (17/19 + 10/19) + 7/19

= 15/19 . 27/19 + 7/19

= 405/361 + 7/19

= 538/361

a)\(\dfrac{8}{15}< \dfrac{11}{15}< \dfrac{15}{15}< \dfrac{16}{15}< \dfrac{17}{15}< \dfrac{19}{15}\)

b)\(\dfrac{19}{42}< \dfrac{19}{35}< \dfrac{19}{21}< \dfrac{19}{19}< \dfrac{19}{17}\)

c)\(\dfrac{8}{10}< \dfrac{6}{7}< \dfrac{27}{25}< \dfrac{16}{14}\)

a) 8/15 ;  11/15 ; 15/15 ; 16/15 ; 17/15 ; 19/15

b) 19/42 19/35 ; 19/21 ; 19/19 ; 19/17

c) 8/10 ; 6/7 ; 27/25 ; 16/14

\(A=\frac{108}{199}\)x\(\left(\frac{107}{211}+\frac{104}{211}\right)\)

\(A=\frac{108}{199}\)x \(\frac{211}{211}\)

\(A=\frac{108}{199}\)x 1

\(A=\frac{108}{199}\)

\(B=\frac{15}{19}\)x \(\left(\frac{27}{33}+\frac{19}{33}-\frac{13}{33}\right)\)

\(B=\frac{15}{19}\)x\(1\)

\(B=\frac{15}{19}\)

a: =17/25(13/19+6/19)=17/25

b: \(=\dfrac{47}{19}\cdot\dfrac{32}{15}-\dfrac{47}{19}\cdot\dfrac{17}{15}=\dfrac{47}{19}\)

`17/25xx(13/19+6/19)`

`17/25xx1`

`17/25`

`a)17/25xx13/19+17/25xx6/19`

`=17/25xx(13/19+6/19)`

`=17/25xx19/19=17/25`

`b)47/19:15/32-47/19:15/17`

`=47/19xx32/15-47/19xx17/15`

`=47/19xx(32/15-17/15)`

`=47/19xx15/15=47/19`

\(\frac{19}{35}+\frac{15}{19}+\frac{16}{35}+\frac{23}{19}+\frac{4}{15}+\frac{11}{15}\)

\(=\left(\frac{19}{35}+\frac{16}{35}\right)+\left(\frac{15}{19}+\frac{23}{19}\right)+\left(\frac{4}{15}+\frac{11}{15}\right)\)

\(=\frac{35}{35}+\frac{38}{19}+\frac{15}{15}\)

\(=1+2+1\)

\(=4\)

\(\frac{19}{35}+\frac{15}{19}+\frac{16}{35}+\frac{23}{19}+\frac{4}{15}+\frac{11}{15}\) 

\(=\left(\frac{19}{35}+\frac{16}{35}\right)+\left(\frac{15}{19}+\frac{23}{19}\right)+\left(\frac{4}{15}+\frac{11}{15}\right)\)

\(=1+2+1\)

\(=4\)

Học tốt

Ta có : \(A=\frac{19^{30}+15}{19^{31}+15}\)

\(\Rightarrow19A=\frac{19^{31}+285}{19^{31}+15}=\frac{19^{31}+15+270}{19^{31}+15}=1+\frac{270}{19^{31}+15}\)

Lại có \(B=\frac{19^{31}+15}{19^{32}+15}\)

\(\Rightarrow19B=\frac{19^{32}+285}{19^{32}+15}=\frac{19^{32}+15+270}{19^{32}+15}=1+\frac{270}{19^{32}+15}\)

Vì \(\frac{270}{19^{32}+15}< \frac{270}{19^{31}+15}\Rightarrow1+\frac{270}{19^{32}+5}< 1+\frac{270}{19^{31}+15}\Rightarrow19B< 19A\Rightarrow B< A\)

\(\dfrac{15}{19}.\dfrac{24}{23}.\dfrac{-19}{15}.\dfrac{23}{24}=-1\)

\(\dfrac{15}{19}.\dfrac{24}{23}.-\dfrac{19}{15}.\dfrac{23}{24}=\left(-\dfrac{19}{15}.\dfrac{15}{19}\right).\left(\dfrac{24}{23}.\dfrac{23}{24}\right)=-1.1=-1\)

\(\dfrac{15}{19}\cdot\dfrac{24}{23}\cdot\dfrac{-19}{15}\cdot\dfrac{23}{24}\)

\(=\dfrac{15}{19}\cdot\dfrac{-19}{15}\cdot\dfrac{23}{24}\cdot\dfrac{24}{23}\)

\(=-1\cdot1=-1\)