Câu hỏi của Trần Thị Ngát

Question

cho $S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}$ không tính tổng S, hãy chứng minh S không phải 1 số tự nhiên

cho

ST · Accepted Answer

a,Ta có: $\frac{3}{10}=\frac{3}{10};\frac{3}{11}< \frac{3}{10};\frac{3}{12}< \frac{3}{10};\frac{3}{13}< \frac{3}{10};\frac{3}{14}< \frac{3}{10}$

$\Rightarrow S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}=\frac{3}{2}=1,5\left(1\right)$

Lại có: $\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}$

$\Rightarrow S>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\left(2\right)$

Từ (1) và (2) => 1 < S < 1,5

Vậy...

b, $A=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{100}$

$=\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)$

Ta có: $\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}$

$\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(1\right)$

Lại có: $\frac{1}{81}>\frac{1}{100};\frac{1}{82}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}$

$\Rightarrow\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{20}{100}=\frac{1}{5}\left(2\right)$

Từ (1) và (2) => $A>\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$

Vậy...

Câu trả lời:

a,Ta có: $\frac{3}{10}=\frac{3}{10};\frac{3}{11}< \frac{3}{10};\frac{3}{12}< \frac{3}{10};\frac{3}{13}< \frac{3}{10};\frac{3}{14}< \frac{3}{10}$

$\Rightarrow S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}=\frac{3}{2}=1,5\left(1\right)$

Lại có: $\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}$

$\Rightarrow S>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\left(2\right)$

Từ (1) và (2) => 1 < S < 1,5

Vậy...

b, $A=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{100}$

$=\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)$

Ta có: $\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}$

$\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(1\right)$

Lại có: $\frac{1}{81}>\frac{1}{100};\frac{1}{82}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}$

$\Rightarrow\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{20}{100}=\frac{1}{5}\left(2\right)$

Từ (1) và (2) => $A>\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$

Vậy...