ta có quy đồng B ta dc(-9x10^2018-19x10^2019)/(10^2019x10^2018)

tương tự với C ta có (-19x10^2018-9x10^2019)/(10^2019x10^2018)

sau khi quy đồng ta thấy mẫu của B và C giống nhau từ đó ta so sánh tử số của B và C

tử số của B=10^2018x(-9-19x10)=10^2018x-199

C=10^2018x(-19-9x10)=10^2018x-109

ta thấy -199<-109=>B<C (dpcm)

B>A

nha bn

vì sao lại lớn hơn

Ta có thể thấy rõ: \(2019< 10^{2019};2019< 10^{2018}\)

\(\Rightarrow-\frac{9}{2019}>-\frac{9}{10^{2019}};-\frac{19}{2019}>-\frac{19}{10^{2018}}\)(do tử là số âm nên mẫu càng lớn thì cả pso càng nhỏ)

\(\Rightarrow-\frac{9}{2019}+\frac{-19}{2019}>-\frac{9}{10^{2019}}+\frac{-19}{10^{2018}}\)

Hay A>B

Chúc bn hok tốt,nhớ k nha

a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A

                                                       Vậy A > B

b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A

                                                                         Vậy A < B.

                    NHỚ K CHO MK VỚI NHÉ !!!!!!!!

a A lon hon B

a) 

\(9A=\frac{9^{19}+9}{9^{19}+1}=\frac{9^{19}+1+8}{9^{19}+1}=1+\frac{8}{9^{19}+1}\)

\(9A=\frac{9^{20}+9}{9^{20}+1}=\frac{9^{20}+1+8}{9^{20}+1}=1+\frac{8}{9^{20}+1}\)

ta thấy \(9^{19}+1< 9^{20}+1\Rightarrow\frac{8}{9^{19}+1}>\frac{8}{9^{20}+1}\)

\(\Rightarrow9A>9B\Rightarrow A>B\)

ta có :

\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)

\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)

Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)

\(\Rightarrow A< B\)

bằng nhau

a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)

\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)

Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)

b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)

\(\frac{2021}{2020}=1+\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)

c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)

\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)

Đến đây tự so sánh rồi nhé

a, Ta có : \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)

\(\Rightarrow\frac{13}{38}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)

b, Ta có: \(3^{301}>3^{300}=\left(3^3\right)^{100}=27^{100}\left(1\right)\)

               \(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\left(2\right)\)

 Do \(25^{100}< 27^{100}\Rightarrow5^{200}< 3^{300}\)\(\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow5^{199}< 5^{200}< 3^{300}< 3^{301}\Rightarrow5^{199}< 3^{301}\)

c, Ta có: \(\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)

               \(\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)

Do \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)