1.

a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{6}{2}.\frac{10}{39}\)

\(=\frac{10}{13}\)

b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)

\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)

\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)

\(=\frac{3}{2}.\frac{5}{28}\)

\(=\frac{15}{56}\)

\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

\(a.\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)

\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)

\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=3.\frac{10}{39}\)

\(=\frac{10}{13}\)

Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)

\(\Rightarrow2017A>2017B\Rightarrow A>B\)

Vậy...

Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)

Hay \(2017A>2017B\)nên \(A>B\)

Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)

đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\); \(B=\frac{2017^{100}+1}{2017^{101}+1}\)

Ta có : \(2017A=\frac{2017.\left(2017^{99}+1\right)}{2017^{100}+1}=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(2017B=\frac{2017.\left(2017^{100}+1\right)}{2017^{101}+1}=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

Vì \(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\Leftrightarrow10A>10B\Rightarrow A>B\)

2016^100+2016^99>2017^100

cái phép tính 1 lớn hơn

2016^100+2016^99= 2016^99.2016^1+2016^99.1=2016^99.(2016+1)=2016^99.2017

2017^100= 2017^99.2017

Do 2016^99.2017< 2017^99.2017

Nên 2016^100+2016^99< 2017^100

a) \(\dfrac{122}{123}\) và \(\dfrac{10}{11}\)

\(1-\dfrac{122}{123}=\dfrac{1}{123}\)              

\(1-\dfrac{10}{11}=\dfrac{1}{11}\)

Vì \(\dfrac{1}{123}< \dfrac{1}{11}\) nên ⇒ \(\dfrac{122}{123}< \dfrac{10}{11}\) 

b) \(\dfrac{16}{12}\) và \(\dfrac{99}{100}\)

\(\dfrac{16}{12}>1\) và \(\dfrac{99}{100}< 1\)

⇒ \(\dfrac{16}{12}>\dfrac{99}{100}\)

c) \(\dfrac{35}{70}\) và \(\dfrac{6}{11}\)

\(\dfrac{35}{70}=\dfrac{1}{2}\) = \(\dfrac{6}{12}\)

Vì \(\dfrac{6}{12}< \dfrac{6}{11}\) nên ⇒ \(\dfrac{35}{70}< \dfrac{6}{11}\)

vào câu hỏi tương tự yk 

quách anh thư mk vào r mà ko có

Công thức: \(\left(a^n\right)^m=a^{n.m}\)

Ta có: \(A=\left(100^{99}+99^{99}\right)^{100}=199^{99.99.100}\)

\(B=\left(100^{100}+99^{100}\right)^{99}=199^{100.100.99}\)

Vì 99.99.100 < 100.100.99 nên \(199^{99.99.100}< 199^{100.100.99}\)

Vậy A < B

a) 16¹⁹ và 8²⁵ 

Ta có :

16¹⁹ = ( 2⁴ )¹⁹ = 2⁷⁶

8²⁵ = ( 2³ )²⁵ = 2⁷⁵

Vì 2⁷⁶ > 2⁷⁵ Nên 16¹⁹ > 8²⁵

b) 5³⁶ và 11²⁴

Ta có :

5³⁶ = ( 5³ )¹² = 125¹²

11²⁴ = ( 11² )¹² = 121¹²

Vì 125¹² > 121¹² Nên 5³⁶ > 11²⁴

1.

\(M=3^0+3^1+......+3^{50}.\)

\(\Rightarrow3M=3+3^2+.......+3^{51}\)

\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)

\(\Rightarrow2M=3^{51}-1\)

\(\Rightarrow M=\frac{3^{51}-1}{2}\)

2.

\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

                     \(8^{25}=\left(2^3\right)^5=2^{75}\)

Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)

\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

                      \(11^{24}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)