#)Giải :

\(A=\frac{20^{18}+1}{20^{19}+1}\)và \(B=\frac{20^{17}+1}{20^{18}+1}\)

\(A=\frac{20^{18}+1}{20^{18+1}+1}\)và \(B=\frac{20^{17}+1}{20^{17+1}+1}\)

\(A=\frac{1}{20+1}\)và \(B=\frac{1}{20+1}\)

\(A=\frac{1}{21}\)và \(B=\frac{1}{21}\)

\(\Rightarrow A=B\)

       #~Will~be~Pens~#

A>20¹⁸ +1+19/20¹⁹ +1+19

A>20¹⁸+20/20¹⁹+20

A>20(20¹⁷+1)/20(20¹⁸+1)

A>20¹⁷+1/20¹⁸+1

=>A>B

Chúc bạn học tốt

\(A=\frac{20^{18}+1}{20^{19}+1}\)

\(A< \frac{20^{18}+1+19}{20^{19}+1+19}=\frac{20^{18}+20}{20^{19}+20}=\frac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}=B\)

\(\Rightarrow A< B\)

                                     #Louis

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\left(19SH\right)\)

\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{20}>\frac{19}{20}\)

Vậy ................

Đặt \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\) ta có : 

\(A>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)

Do có \(20-2+1=19\) phân số \(\frac{1}{20}\) nên : 

\(A>19.\frac{1}{20}=\frac{19}{20}\)

Vậy \(A>\frac{19}{20}\)

Chúc bạn học tốt ~ 

Giúp với

Chứng minh nếu a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)

Do a/b < 1 => a < b

=> am < bm

=> am + ab < bm + ab

=> a.(b+m) < b.(a+m)

=> a/b < a+m/b+m

Áp dụng điều trên ta có: B = 10²⁰ + 1/ 10²¹ + 1 < 1

=> B < 10²⁰ + 1 + 9/10²¹ + 1 + 9

=> B < 10²⁰ + 10/10²¹ + 10

=> B < 10.(10¹⁹ + 1)/10.(10²⁰ + 1)

=> B < 10¹⁹+1/10²⁰+1 = A

=> B < A

b) n + 1 chia hết cho n - 2

=> n - 2 + 3 chia hết cho n - 2

Do n - 2 chia hết cho n - 2

=> 3 chia hết cho n - 2

=> n - 2 thuộc { 1 ; -1 ; 3 ; -3}

=> n thuộc { 3 ; 1 ; 5 ; -1}

Vậy n thuộc { 3 ; 1 ; 5 ; -1}

ehfhthjjm458855 ngu

Ta có:

\(20A=\frac{20\left(20^{19}+1\right)}{20^{20}+1}=\frac{20^{20}+20}{20^{20}+1}=\frac{20^{20}+1+19}{20^{20}+1}=\frac{20^{20}+1}{20^{20}+1}+\frac{19}{20^{20}+1}=1+\frac{19}{20^{20}+1}\)

\(20B=\frac{20\left(20^{20}+1\right)}{20^{21}+1}=\frac{20^{21}+20}{20^{21}+1}=\frac{20^{21}+1+19}{20^{21}+1}=\frac{20^{21}+1}{20^{21}+1}+\frac{19}{20^{21}+1}=1+\frac{19}{20^{21}+1}\)

Vì 20²⁰+1<20²¹+1

\(\Rightarrow\frac{19}{20^{20}+1}>\frac{19}{20^{21}+1}\)

\(\Rightarrow1+\frac{19}{20^{20}+1}>1+\frac{19}{20^{21}+1}\)

\(\Rightarrow20A>20B\)

\(\Rightarrow A>B\)

Ta có: \(20A=\dfrac{20^{19}+20}{20^{19}+1}=1+\dfrac{19}{20^{19}+1}\)

\(20B=\dfrac{20^{18}+20}{20^{18}+1}=1+\dfrac{19}{20^{18}+1}\)

Vì \(\dfrac{19}{20^{19}+1}< \dfrac{19}{20^{18}+1}\Rightarrow1+\dfrac{19}{20^{19}+1}< 1+\dfrac{19}{20^{18}+1}\)

\(\Rightarrow20A< 20B\Rightarrow A< B\)

Vậy A < B

Ta có: \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)(a \(\in\) N và b,c,d \(\in\) N*)

Áp dụng kiến thức đó, ta được:

A = \(\dfrac{20^{18}+1}{20^{19}+1}\) <\(\dfrac{20^{18}+1+19}{20^{19}+1+19}\)= \(\dfrac{20^{18}+20}{20^{19}+20}\) = \(\dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}\)

= \(\dfrac{20^{17}+1}{20^{18}+1}\) = B

Vậy A < B

Ta có: \(\dfrac{20^{18}+1}{20^{19}+1}< \dfrac{20^{18}+1+19}{20^{19}+1+19}\)

\(\Rightarrow A< \dfrac{20^{18}+20}{20^{19}+20}\)

\(\Rightarrow A< \dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}\)

\(\Rightarrow A< \dfrac{20^{17}+1}{20^{18}+1}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

Ta có : 

\(A=\frac{1}{2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{2}\left(\frac{189}{380}\right)=\frac{189}{760}< \frac{1}{4}\)

Ta có: \(A=\frac{1}{2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{18\times19\times20}\)

              \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{18\times19}-\frac{1}{19\times20}\right)\)

               \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{19\times20}\right)\)

                 \(=\frac{1}{2}\times\frac{1}{1\times2}-\frac{1}{2}\times\frac{1}{19\times20}\)

                   \(=\frac{1}{4}-\frac{1}{2}\times\frac{1}{19\times20}< \frac{1}{4}\)

Vậy A < 1/4

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!