\(B=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{30}\cdot20=\frac{2}{3}\)

\(B< \frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{1}{10}\cdot20=2\)

\(\Rightarrow\frac{99}{100}< \frac{2}{3}< B< 2\)

1/

\(10A=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 1< 10B$

$\Rightarrow A< B$

2/

\(C=\frac{10^{99}+5}{10^{99}-8}=1+\frac{13}{10^{99}-8}\)

\(D=\frac{10^{100}+6}{10^{100}-4}=1+\frac{10}{10^{100}-4}\)

So sánh \(\frac{13}{10^{99}-8}=\frac{130}{10^{100}-80}> \frac{130}{10^{100}-4}> \frac{10}{100^{100}-4}\)

$\Rightarrow 1+\frac{13}{10^{99}-8}> 1+\frac{10}{100^{10}-4}$

$\Rightarrow C> D$

Câu hỏi của Nguyễn Văn Bình - Toán lớp 6 - Học toán với OnlineMath

1/100<1

ch h t,nhớ mk cho m nha

a: Ta có: \(A=\frac59+\left(-\frac57\right)+\left(-\frac{20}{48}\right)+\frac{8}{12}+\left(-\frac{21}{48}\right)\)

\(=\frac59-\frac57-\frac{41}{48}+\frac{32}{48}\)

\(=\frac{35-45}{63}-\frac{9}{48}=\frac{-10}{63}-\frac{3}{16}=\frac{-160-189}{63\cdot16}=\frac{-349}{1008}\)

b: \(B=\left(-\frac59\right)+\frac{8}{15}+\left(-\frac{2}{11}\right)+\left(\frac{4}{-9}\right)+\frac{2}{45}\)

\(=\left(-\frac59-\frac49\right)+\frac{8}{15}+\frac{2}{45}-\frac{2}{11}\)

\(=-1-\frac{2}{11}+\frac{24}{45}+\frac{2}{45}=-\frac{13}{11}+\frac{26}{45}=\frac{-13\cdot45+26\cdot11}{11\cdot45}=\frac{-299}{495}\)

c: \(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\ldots;\frac{1}{20}=\frac{1}{20}\)

Do đó: \(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}\)

=>S>10/20

=>S>1/2

Sửa đề: \(S=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)

Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)

=>\(S<\frac56\)

P=1/10+1/11+...+1/100=1/10+(1/11+1/12+...+1/50)+(1/51+1/52+...+1/100)

Đặt A = 1/11+1/12+1/13+...+1/50

A có (50-11):1+1=40(số hạng)

Lại có: 1/11>1/12>...>1/50

=>1/11+1/12+1/13+...+1/50>1/50+1/50+...+1/50(40 số hạng)

=>A>4/5

Đặt B =1/51+1/52+...+1/100

B có (100-51):1+1=50 (số hạng)

Lại có : 1/51>1/52>...>1/100

=>1/51+1/52+1/53+...+1/100>1/100+1/100+...+1/100(50 số hạng)

=>B>1/2

=>P>1/10+4/5+1/2

=>P>14/10

=>P>1

Vậy P>1

kb đc 0

2 câu đầu tôi làm đc

C>1   vì c>1

a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)

Vậy A > 1/2

b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Vậy B > 1/2

c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)

Vậy C > 1