\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{19}{9^2.10^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+......+\frac{10^2-9^2}{9^2.10^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+.....+\frac{1}{9^2}-\frac{1}{10^2}\)

\(=\frac{1}{1^2}-\frac{1}{10^2}=1-\frac{1}{10^2}<1\left(đpcm\right)\)

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.

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

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

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

So so hang trong day la:

(50-21) : 1 + 1 = 30

Gia su = 1/50 het thi tong la:

30 x 1/50 = 3/5

Vi 1/50 la so nho nhat nen tong se lon hon 3/5

Gia su tat ca deu la 1/21

Tong la:

1/21 x 30 = 30/21 = 10/7

So sanh 10/7 va 3/2, ta thay 3/2 lon hon 10/7 la 1/14 don vi

Vi 1/21 la so lon nhat nen tong se be hon 3/2

ta có:A= 1/21 + 1/22 + ... + 1/50 > 1/50 +1/50 +...+1/50=1/50 x 30 = 3/5

=> A > 3/5

lại có: A = 1/21 + 1/22 + ... + 1/50 < 1/20 + 1/20  + ... +1/20= 1/20 x 30 = 3/2

=> A <3/2

cách này là cách nhanh nhất

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>\(\frac{9}{10}\)