Vì N<1

=> N= 20^31+2/20^32+2

<20^31+2+38/ 20^32+2+38

=20^31+40/ 20^32+40

=20.(20^30+2) / 20.(20^31+2)

=20^30+2 / 20^32+2 = M

Vậy N<M

\(N=\frac{20^{31}+2}{20^{32}+2}=\frac{20^{31}+2+18}{20^{32}+2+18}=\frac{20^{31}+20}{20^{32}+20}=\frac{10.\left(20^{30}+2\right)}{10.\left(20^{31}+2\right)}\)\(=M\)

\(\Rightarrow M=N\)
 

a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)

b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)

\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)

Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

c,  Câu hỏi của truong nguyen kim 

a) Ta có: \(A=\frac{2^{2017}}{2^{2017}}+\frac{2^{2016}}{2^{2017}}+\frac{2^{2015}}{2^{2017}}+...+\frac{2^1}{2^{2017}}+\frac{1}{2^{2017}}\)

\(=\frac{1+2^1+2^2+...+2^{2016}+2^{2017}}{2^{2017}}\)

Đặt: B=\(1+2^1+2^2+...+2^{2017}\)

\(\Leftrightarrow2B=2^1+2^2+2^3+....+2^{2017}+2^{2018}\)

\(\Leftrightarrow2B-B=2^{2018}-1\)

\(\Leftrightarrow B=2^{2018}-1\)

\(\Rightarrow A=\frac{B}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)

Mik chỉ biết làm phần a thôi

b/ Sử dụng quy tắc: \(\frac{a+c}{b+c}< \frac{a}{b}\) với \(\left\{{}\begin{matrix}a;b;c>0\\a>b\end{matrix}\right.\)

\(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)

\(\Rightarrow B>A\)

Ta có a/b >1 => a/b > a+n/b+n(a, b,n $\in$∈ N*)               

B = 20¹⁰-1/20¹⁰-3 > 1 nên B = 20¹⁰-1/20¹⁰-3 > 20¹⁰-1+2/20¹⁰-3+2  

   = 20¹⁰+1/ 20¹⁰-1 = A

Vay \(A=\frac{2^{10}+1}{20^{10}-1}<\frac{20^{10}-1}{20^{10}-3}\) 

 a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)

Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)

Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).

b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)

Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)

Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)

\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)

\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)

Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).

d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)

Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)

Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).

Rút gọn biểu thức trên nha.

\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)

\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)

vậy M=20