11¹³+1/ 11¹⁴+1 = 11¹⁴+1/11¹⁵+1

Với n > 0 Ta có:

\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)

\(=\sqrt{n+1}+\sqrt{n}\)

\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)

\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)

\(\sqrt{16}+\sqrt{9}=3+4=7\)

\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)

\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)

\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}\)

Bài làm

Ta có: 

\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)

=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)

hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)

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

Do đó: \(S=\frac{1}{2}\)

# Chúc bạn học tốt #

Ta có các phân số : \(\frac{1}{11};\frac{1}{12};\frac{1}{13};\frac{1}{14};\frac{1}{15};\frac{1}{16};\frac{1}{17};\frac{1}{18};\frac{1}{19}>\frac{1}{20}\)

Do đó : \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)có 10 phân số \(\frac{1}{20}\)

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{10}{20}\)

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{2}\)

Vậy : \(S>\frac{1}{2}\)

\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)

\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)

\(A\)\(=\)\(\frac{2}{45}\)

\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)

Sau đó nhân phân phối ra là xong nhé bạn 

Có : 10A = 10^15-10/10^15-11 = (10^15-11)+1/10^15-11 = 1 + 1/10^15-11

10B = 10^15+10/10^15+9 = (10^15+9)+1/10^15+9 = 1 + 1/10^15+9

Vì 10^15-11 < 10^15-9 => 1/10^15-11 > 1/10^15+9 => 10A > 10B

=> A < B

k mk nha

Bài 1 :

Đặt \(A=\frac{11^{13}+1}{11^{14}+1}\) và \(B=\frac{11^{14}+1}{11^{15}+1}\)

Có : \(A=\frac{11^{13}+1}{11^{14}+1}\)

\(\Rightarrow11A=\frac{11^{14}+11}{11^{14}+1}=\frac{11^{14}+1+10}{11^{14}+1}=1+\frac{10}{11^{14}+1}\)

Lại có : \(B=\frac{11^{14}+1}{11^{15}+1}\)

\(\Rightarrow11B=\frac{11^{15}+11}{11^{15}+1}=\frac{11^{15}+1+10}{11^{15}+1}=1+\frac{10}{11^{15}+1}\)

Vì 11¹⁴+1<11¹⁵+1

\(\Rightarrow\frac{10}{11^{14}+1}>\frac{10}{11^{15}+1}\Rightarrow1+\frac{10}{11^{14}+1}>1+\frac{10}{11^{15}+1}\Rightarrow11A>11B\Rightarrow A>B\)

Vậy A>B.

Bài 2 :

a) Gọi (n+1,2n+3) là d  (d là số tự nhiên khác 0)

\(\Rightarrow\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}}\)

\(\Rightarrow\left(2n+3\right)-\left(n+1\right)⋮d\)

\(\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

nên (n+1,2n+3) là 1

\(\Rightarrow\frac{n+1}{2n+3}\) là phân số tối giản(đpcm)

b) Gọi (12n+1,30n+2) là d  (d là số tự nhiên khác 0)

\(\Rightarrow\hept{\begin{cases}12n+1⋮d\\30n+2⋮d\end{cases}}\)

\(\Rightarrow\left(12n+1\right)-\left(30n+2\right)⋮d\)

\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

nên (12n+1,30n+2) là 1

\(\Rightarrow\)Phân số \(\frac{12n+1}{30n+2}\)tối giản(đpcm)

c và d tương tự

A><=1

k mình nha

Gợi ý: Rút gọn 2 ps, quy đồng rồi so sánh.