Y Ribi Nkok Ngok Lê Nguyễn Ngọc Nhi Lê Anh Duy Nguyễn Thị Diễm Quỳnh trần thị diệu linh kudo shinichi Nguyen Giang Thủy Tiên Nguyễn Việt Lâm

gọi A=1/21+1/22+1/23+...+1/40

chia A thành 2 nhóm A1 và A2( A1+A2=A)

ta có A1=1/21+1/22+1/23+...+1/30>1/30+1/30+1/30+...+1/30(có 10 phân số 1/30)

A1>10/30=1/3(1)

ta có A2=1/31+1/32+1/33+...+1/40>1/40+1/40+1/40+...+1/40(có 10 phân số 1/40)

A2>10/40=1/4(2)

từ (1)và (2) suy ra

A1+A2>1/3+1/4

A>7/12(3)

ta có A1=1/21+1/22+1/23+...+1/20<1/20+1/20+1/20+...+1/20(có 10 phân số 1/20)

A1<10/20=1/2(4)

ta có A2=1/31+1/32+1/33+...+1/40<1/30+1/30+1/30+...+1/30(có 10 phân số 1/30)

A2<10/30=1/3(5)

từ (4)và (5) suy ra

A1+A2<1/2+1/3

A<5/6(6)

từ (3),(6) suy ra 7/12<1/21+1/22+1/23+...+1/40<5/6

cái A1+1/21+1/22+1/23+1/24+1/25+...+1/30<1/20+1/20+1/20+1/20+...+1/20 nhé

a)đặt B=1/2.3+1/3.4+...+1/99.100

=1/1.2+1/2.3+1/3.4+...+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100<1 (1)

Mà 1<2(2)

A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)

từ (1),(2),(3) =>A<2

b,c tự làm

Thế mà ko biết làm

Thế mà ko biết làm

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

tách bất đẳng thức trên ta có \(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)gọi biều thức này là A

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

\(A=\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)

\(\Rightarrow A>20.\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+40.\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)nhân vế trái vs 20 vế phải 40

\(\Rightarrow A>20.\left(\frac{1}{20}-\frac{1}{40}\right)+40.\left(\frac{1}{40}-\frac{1}{60}\right)\)

\(\Rightarrow A>\frac{5}{6}>\frac{11}{5}\left(1\right)\)

ta có \(A< 40.\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+60.\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)

\(\Rightarrow A< 40.\left(\frac{1}{20}-\frac{1}{40}\right)+60.\left(\frac{1}{40}-\frac{1}{60}\right)\)

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

từ (1) và (2)

\(\Rightarrow\frac{11}{15}< A< \frac{3}{2}\)

\(\Rightarrow\frac{11}{15}< \text{​​}\text{​​}\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+..+\frac{1}{60}< \frac{3}{2}\)(ĐPCM)

Đáp án là mình chứng minh được.