Ta có \(A=\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{39}\right)+\left(\frac{1}{40}+\frac{1}{41}+...+\frac{1}{59}\right)\)

\(A< \left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

\(A< \frac{20}{20}+\frac{20}{40}\)

\(A< \frac{3}{2}\)

Câu hỏi của Hoàng Đỗ Việt - Toán lớp 6 | Học trực tuyến

Bài 1 :

Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)

Do đó :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)

Mặt khác :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)

\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)

Do đó :

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)

Từ (1 ) và (2) ta suy ra điều phải chứng minh

Bài 2 : 

Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)

MỘT MẶT ,TA CÓ THỂ VIẾT

\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)

\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)

Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)

Từ (1) và (2 ) ta kết luận \(3< S< 6\)

Chúc bạn học tốt ( -_- )

a) Đặt \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}\)

Ta có:

\(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

+ Vì \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};...;\frac{1}{40}=\frac{1}{40}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}.\)

+ Vì \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{60}=\frac{1}{60}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( 20 phân số \(\frac{1}{60}\)) \(=20.\frac{1}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}=\frac{75}{90}>\frac{66}{90}=\frac{11}{15}\)

\(\Rightarrow A>\frac{11}{15}\left(1\right)\)

Lại có: 

+ Vì \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...;\frac{1}{40}< \frac{1}{20}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( 20 phân số \(\frac{1}{20}\)) \(=20.\frac{1}{20}=1\)

+ Vì \(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{60}< \frac{1}{40}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}\)

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

\(\Rightarrow A< \frac{3}{2}\left(2\right)\)

Từ\(\left(1\right)\);\(\left(2\right)\) \(\Rightarrow\frac{11}{15}< A< \frac{3}{2}\left(đpcm\right).\)

b) Đặt \(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)

Ta có:

\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)\)\(+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)

+ \(1=1\)

+\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)

+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)

Tương tự ta được:

+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< 1\)

+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< 1\)

\(\Rightarrow A< 1+1+1+1+1+1=6\left(1\right)\)

Lại có:

+\(1=1\)

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

+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{4}{7}\)

+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}>\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}=\frac{8}{15}\)

Tương tự, ta được:

+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}>\frac{16}{31}\)

+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< \frac{32}{63}\)

\(\Rightarrow A>1+\frac{2}{3}+\frac{4}{7}+\frac{8}{15}+\frac{16}{31}+\frac{32}{63}\)\(=1+\frac{18}{15}+\frac{64}{63}+\frac{16}{31}>1+\frac{15}{15}+\frac{63}{63}=3\left(2\right)\)

Từ\(\left(1\right)\)và \(\left(2\right)\Rightarrow3< A< 6\left(đpcm\right).\)

S = \(\frac{1}{20}+\frac{1}{21}...+\frac{1}{199}+\frac{1}{200}\)  ( có 181 phân số )

=> S > \(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}+\frac{1}{200}\)

=> S > \(\frac{1}{200}.181\)

=> S > \(\frac{181}{200}\)> \(\frac{180}{200}\)= \(\frac{9}{10}\)

Vậy S > 9 / 10

GIÚP NHA , AI LÀM ĐƯƠC 1 NGÀY TK 3TK

S = \(\frac{1}{20}\)+ \(\frac{1}{21}\)+ ....+\(\frac{1}{200}\)có 181 p/s

mà \(\frac{1}{20}\)>\(\frac{1}{200}\)

.............

    \(\frac{1}{199}\)>\(\frac{1}{200}\)

    \(\frac{1}{200}\)=\(\frac{1}{200}\)

nên  ta có     S > \(\frac{1}{200}\)+ \(\frac{1}{200}\)+..... có 181 phân số \(\frac{1}{200}\)

vậy \(\frac{1}{200}\)*181=\(\frac{181}{200}\)mà \(\frac{181}{200}\)>\(\frac{9}{10}\)mà \(\frac{1}{20}\)+......+\(\frac{1}{200}\)(có 181 số)>\(\frac{1}{200}\)+\(\frac{1}{200}\)(có 181 p/s \(\frac{1}{200}\))>\(\frac{9}{10}\)

Vậy ==> S>\(\frac{9}{10}\)

Đặt \(C=\frac{1}{21}+\frac{1}{22}+....+\frac{1}{60}=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};....\frac{1}{39}>\frac{1}{40}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+....+\frac{1}{39}+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\) 

\(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...\frac{1}{59}>\frac{1}{60}\)

 \(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{1}{60}.20=\frac{1}{3}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{11}{15}\)

Vậy \(C>\frac{11}{15}\) (1)

Lại có: \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...\frac{1}{40}< \frac{1}{20}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+....+\frac{1}{20}=\frac{1}{20}.20=1\)

\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...\frac{1}{60}< \frac{1}{40}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{1}{2}+1=\frac{3}{2}\)

Vậy \(C< \frac{3}{2}\) (2)

Từ (1) và (2) suy ra \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{3}{2}\)

Đặt  \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)

=> \(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

Đặt A < (1/40+.....+1/40)+(1/60+1/60+...+1/60)

=>A<1/2+1/3=5/6<3/2

lớn hơn 11/15 cũng tương tự thôi bạn tự làm sẽ thú vị hơn đấy

k minh nha

Thank you

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