\(\frac{-13}{14}>\frac{-15}{16}\)

Ta có:

\(A=\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\frac{10^{16}+10}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

Ta so sánh \(10A\) và \(10B\)

Có: 

\(10A:\) Mẫu - tử = 9

\(10B:\) Mẫu - tử = 9

Lại có:

 \(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)

\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)

Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Theo bải ra ta có:

A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)= \(\frac{10.10^{15}+1.10}{10^{16}+1}\)

                                      = \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)= \(1+\frac{9}{10^{16}+1}\)

B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)

                                       = \(\frac{10.10^{16}+10}{10^{17}+1}\)= \(\frac{10^{17}+1+9}{10^{17}+1}\)= \(1+\frac{9}{10^{17}+1}\)

Vì 1=1 mà \(\frac{9}{10^{16}+1}\)>   \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B

Vậy A>B.

giúp mình với mai phải nộp rồi

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

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

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

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

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

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

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 : \(S>\frac{1}{2}\)

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) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)

Đặt \(B=1+7+7^2+...+7^{14}\)

\(\Rightarrow7B=7+7^2+...+7^{15}\)

\(\Rightarrow7B-B=6B=7^{15}-1\)

\(\Rightarrow B=\frac{7^{15}-1}{6}\)

\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)

Tự làm tiếp nha

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