f,F=3. (1/2 .3 + 1/3.4 +...+ 1/99.100)

    = 3. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100

    = 3. (1/2 - 1/100)

    = 3. 49/100

    = 147/100

g, G = 5/3. (3/1.4 + 3/4.7 +...+ 3/61.64)

        = 5/3 . (1 - 1/4 + 1/4 - 1/7 +...+ 1/61 - 164

        = 5/3 . (1-1/64)

        = 5/3 . 63/64

        = 105/64

f,    \(F=\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

\(\Leftrightarrow F=3\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Leftrightarrow F=3\left(\frac{49}{100}\right)=\frac{147}{100}\)

g,    \(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)

\(\Leftrightarrow G=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{61.64}\right)\)

\(\Leftrightarrow G=5.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)

\(\Leftrightarrow G=\frac{5}{3}\left(1-\frac{1}{64}\right)\)

\(\Leftrightarrow G=\frac{5}{3}.\frac{63}{64}=\frac{105}{64}\)

\(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)

\(\Rightarrow G=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{61.64}\right)\)

\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{61}-\frac{1}{64}\right)\)

\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{64}\right)=\frac{5}{3}.\frac{63}{64}\)

\(\Rightarrow G=\frac{5.63}{3.64}=\frac{5.21.3}{3.64}=\frac{5.21}{64}=\frac{105}{64}\)

a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)

\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)

\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)

Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)

\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)

\(2A=\frac{12}{3}-\frac{12}{99}\)

\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)

1.5/1-5/4+5/4-5/7+5/7-5/9 +....+5/27-5/30

=5/1-5/30

=145/30=29/6.

Làm từng phần nha bạn

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)

Đặt \(A+x=\frac{299}{301}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)

\(A=1-\frac{1}{301}\)

\(A=\frac{300}{301}\)

=> \(\frac{300}{301}+x=\frac{299}{301}\)

\(x=\frac{299-300}{301}\)

\(x=-\frac{1}{301}\)

\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)

\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=\frac{303}{304}\)

\(A=\frac{505}{304}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{100.103}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

=5/3.(3/1.4+3/4.7+3/7.10+...+3/100.103)

=5/3.(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)

=5/3.(1-1/103)=5/3.102/103=170/103

                                       đáp số : 170/103

ta nhân 3 cả hai vế, được : 

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)

hay 

\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)

\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)

5/1.4 + 5/4.7 + 5/7.10 + ... + 5/97.100

= 5/3 . (3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100)

= 5/3 . (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100)

= 5/3 . ( 1 - 1/100)

= 5/3 . 99/100

= 33/20

a) A = 3/10.12 + 3/12.14 + ... + 3/998.1000

2/3.A = 2/10.12 + 2/12.14 + ... + 2/998.1000

2/3.A = 1/10 - 1/12 + 1/12 - 1/14 + ... + 1/998 - 1/1000

2/3.A = 1/10 - 1/1000

2/3.A = 99/1000

A = 99/1000 : 2/3

A = 99/1000 . 3/2

A = 297/2000

b) B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/22.25

3/2.B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/22.25

3/2.B = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/22 - 1/25

3/2.B = 1 - 1/25

3/2.B = 24/25

B = 24/25 : 3/2

B = 24/25 . 2/3

B = 16/25

Ủng hộ mk nha ^_-

a) Ta có: \(A=\frac{3}{10.12}+\frac{3}{12.14}+....+\frac{3}{998.1000}.\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10.12}+\frac{1}{12.14}+...+\frac{1}{998.1000}\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{998}-\frac{1}{1000}\)

 \(\Rightarrow\frac{2}{3}A=\frac{1}{10}-\frac{1}{1000}=\frac{99}{1000}\)

 \(\Rightarrow A=\frac{99}{1000}:\frac{2}{3}=\frac{297}{2000}\)

Ta có: \(B=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{22.25}\)

\(\Rightarrow\frac{3}{2}B=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{22}-\frac{1}{25}\)

\(\Rightarrow\frac{3}{2}B=1-\frac{1}{25}=\frac{24}{25}\)

\(\Rightarrow B=\frac{24}{25}:\frac{3}{2}=\frac{16}{25}\)

a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)