\(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)

\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)

Nhận thấy 19¹⁷ + 19 > 19¹⁶ + 19

=> \(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)

=> \(-\frac{18}{19^{17}+19}>-\frac{18}{19^{16}+19}\)

=> \(1-\frac{18}{19^{17}+19}>1-\frac{18}{19^{16}+19}\)

=> \(\frac{x}{19}>\frac{y}{19}\)

=> x > y

Vậy x > y

Ta có : \(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)

\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)

Vì\(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)\(\Rightarrow\frac{x}{19}>\frac{y}{19}\)

mà \(x,y>0\)

\(\Rightarrow x>y\)

\(S=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(10 số hạng) 

\(=10.\frac{1}{20}=\frac{1}{2}\).

Vậy \(S>\frac{1}{2}\).

ta thấy: 1/11;1/12;1/13;...;1/19;1/20 đều >1/20

=>1/11+1/12+...1/19+1/20>1/20+1/20...+1/20

1/11+1/12+...1/19+1/20>10/20

1/11+1/12+...1/19+1/20>1/2 vậy S>1/2

Ta thấy mỗi số hạng của tổng đều bé hơn 1/10

=>S<\(\frac{1}{10}.10=1\)

=>S<1

S = 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/19 + 1/20

S < 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10

S < 10 × 1/10

S < 1

M = \(15.\left(\frac{1}{15.16}+\frac{1}{16.17}+...+\frac{1}{19.20}\right)\)
= \(15.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+...+\frac{1}{19}-\frac{1}{20}\right)\)
= \(15.\left(\frac{1}{15}-\frac{1}{20}\right)\)
= \(15.\frac{1}{60}\)= \(\frac{1}{4}\)\(< \frac{1}{3}\)
(=) \(M< \frac{1}{3}\)\(\left(đpcm\right)\)

Ta có: \(M=\frac{15}{15.16}+\frac{15}{16.17}+\frac{15}{17.18}+\frac{15}{18.19}+\frac{15}{19.20}\)

\(\Rightarrow M=15.\left(\frac{1}{15.16}+\frac{1}{16.17}+\frac{1}{17.18}+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(\Rightarrow M=15.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+\frac{1}{17}-\frac{1}{18}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(\Rightarrow M=15.\left(\frac{1}{15}-\frac{1}{20}\right)\)

\(\Rightarrow M=15.\frac{1}{60}=\frac{1}{4}\)

Ta thấy: \(\frac{1}{4}< \frac{1}{3}\Rightarrow M< \frac{1}{3}\)

Vậy \(M< \frac{1}{3}\)

Chúc bạn học tốt!

\(\Rightarrow M=15.\left(\frac{1}{15.16}+\frac{1}{16.17}+...+\frac{1}{19.20}\right)\)

\(M=15.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(M=15.\left(\frac{1}{15}+\left(-\frac{1}{16}+\frac{1}{16}\right)+\left(-\frac{1}{17}+\frac{1}{17}\right)+...+\left(\frac{-1}{19}+\frac{1}{19}\right)-\frac{1}{20}\right)\)

\(M=15.\left(\frac{1}{15}+0+0+0+...+0-\frac{1}{20}\right)\)

\(M=15.\left(\frac{1}{15}-\frac{1}{20}\right)=\frac{15}{60}\)

Mà \(\frac{1}{3}=\frac{20}{60}\)

\(\Rightarrow M< \frac{1}{3}\)

Vậy : \(M< \frac{1}{3}\)

S < 1/2

Ai trả lời mik sẽ k cho người đó bất kể số lượng người

mik viết nhầm mik sẽ k cho người đó

S > 1/2

 Chúc bn học tốt!

có  \(\frac{1}{20}\) bé nhất suy ra

"có 10 số hạng "\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+......+\frac{1}{20}>\frac{1}{20}.10\)

\(VT>\frac{10}{20}=\frac{1}{2}\)

vì 1/11>1/20

     1/12>1/20...

     1/13>1/20

nên 1/11+1/12+,,,,+1/20>1/20+1/20+,,,,+1/20=10/20=1/2(rút gọn

                                    10 số 1/20

vậy S>1/2