Áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

VT=\(x-\left(\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-...\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{99}-\frac{1}{100}\right)\right)\)

=\(x-\frac{1}{100}\)

Dễ dàng tìm được 

\(x-\frac{1}{100}=\frac{1}{100}\) 

\(x=\frac{1}{50}\)

...

= 1/2-1/3+1/3-1/4+...+ 1/19-1/20

= 1/2-1/20

=9/20

có phải như thế này ko bn

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)

A = \(\frac{9}{20}\)

\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)

B = \(-\frac{99}{100}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)

Tương tự với ý B

\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)

MÌNH NGHĨ LÀ A< B

Ta có : 

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=\)\(1-\frac{1}{100}\)

\(=\)\(\frac{99}{100}\)

Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}=\frac{99}{100}\)

Chúc bạn học tốt ~

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

                                                              \(=1-\frac{1}{100}=\frac{99}{100}\)

ĐÚNG 100%

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Làm tiếp

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{99}-\frac{1}{100}\)

A=\(1-\frac{1}{100}\)

A=\(\frac{100}{100}-\frac{1}{100}\)

A=\(\frac{99}{100}\)

A= 2-1/1.2 + 3-2/2.3 + 4-3/3.4 +...+ 99-98/98.99 + 100-99/99.100

A= 2/1.2 - 1/1.2 + 3/2.3 - 2/2.3 + 4/3.4 - 3/3.4 +...+ 99/98.99 - 98/98.99 + 100/99.100 - 99/99.100

A= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/98 - 1/99 + 1/99 - 1/100

A= 1 - 1/100

A= 99/100

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

Ta có công thức: 

\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}=\frac{1}{n.\left(n+1\right)}\)

Áp dụng công thưc trên ta có

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\text{ CMR : }\frac{7}{12}< A< \frac{5}{6}\)

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{98}\)

\(A=1+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{49}\)

\(A=\frac{1}{51}+...+\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{51.25}=\frac{25}{51}< \frac{25}{30}=\frac{5}{6}\) (đpcm)

Và \(A>25.\frac{1}{75}+25.\frac{1}{100}=\frac{7}{12}\)

Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)

Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)

                =  (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <  1 - 1 / 2 + 1 / 3 = 5 / 6 (2)

Từ (1) và (2) => 7 / 12 < A < 5 / 6

trả lời :

Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)

Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)

                =  (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <  1 - 1 / 2 + 1 / 3 = 5 / 6 (2)

Từ (1) và (2) => 7 / 12 < A < 5 / 6

^HT^