Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)

Vậy A < B

lolllllo

Do \(\dfrac{10^{18}+1}{10^{19}+2}< 1\Rightarrow B< \dfrac{10^{18}+1+9}{10^{19}+1+9}\)

\(\Rightarrow B< \dfrac{10^{18}+10}{10^{19}+10}\)

\(\Rightarrow B< \dfrac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)

\(\Rightarrow B< \dfrac{10^{17}+1}{10^{18}+1}\)

\(\Rightarrow B< A\)

So sánh 2 phân số sau  $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10 

kick dzô chữ xanh là được!! OK

Ta có : 

10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)

         = \(\frac{10^{2012}+10}{10^{2012}+1}\)

         = \(\frac{10^{2012}+1+9}{10^{2012}+1}\)

         = \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)

         = 1 - \(\frac{9}{10^{2012}+1}\)

10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)

          = \(\frac{10^{2013}+10}{10^{2013}+1}\)

          = \(\frac{10^{2013}+1+9}{10^{2013}+1}\)

          = 1 - \(\frac{9}{10^{2013}+1}\)

Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\)  nên 10.A > 10.B

=> A >B 

Vậy ...........

Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

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

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

\(10^{17}

A < B                                                            

Nguyen Huynh Truc Anh

TRA LOI SAI ROI:

\(\left(\frac{1}{10}\right)^{18}>\left(\frac{1}{10}\right)^{19}\)

ta thấy :

\(\frac{10^{18}+4}{10^{19}-1}>\frac{10^8}{10^{19}-1}\)

nhưng

\(\frac{10^8}{10^{19}-5}\)<\(\frac{10^8}{10^{19}-1}\)

=>\(\frac{10^{18}+4}{10^{19}-1}>\frac{10^8}{10^{19}-5}\)

(dạng toán so sánh, dùng một số trung gian)