a)1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72+89/90

=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90

=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)

=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]

=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

=9 – (1 – 1/10) = 9 – 9/10 = 81/10

b)4/3.7 + 4/7.11 + 4/11.15 + 4/15.19 + 4/19.23 + 4/23.27

=4.(4/3.7 + 4/7.11 + ........+ 4/23.27 )

=1.( 1/3.7 + 1/7.11 + ......+ 1/23.27 )

=1.(1/3 - 1/7 + 1/7 - 1/11 +............ + 1/23 - 1/27 )

=1.(1/3 - 1/27 )

=1.(9/27 - 1/27)

=1.8/27

=8/27

c)1/10+1/40+1/88+1/154+1/138+1/340

=1/2.5 + 1/5.8 + 1/11.8 + 1/11.14 + 1/14.17 + 1/17.20

=1/3. (3/2.5  + 3/5.8 + 3/8.11 + 3/11.14 + 3/14.17 + 3/17.20 )

=1/3. ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20 )

=1/3. ( 1/2 - 1/20 )

=1/3. 9/20

=3/20

P/S: CHÚC HOK TỐT !

1.\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{23}-\frac{4}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

2. Đặt \(A=\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+\frac{3}{374}+\frac{3}{594}+\frac{3}{864}\)

\(\Rightarrow A=\frac{3}{2.7}+\frac{3}{7.12}+...+\frac{3}{27.32}\)

\(\Rightarrow5A=3.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{27.32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\frac{15}{32}=\frac{45}{32}\Rightarrow A=\frac{45}{32}:5=\frac{9}{32}\)

3. Đặt \(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\)

\(\Rightarrow3S=\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)

Câu 1:

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

Câu 1:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

Áp dụng tính chất \(\frac{b}{a\left[a+b\right]}=\frac{1}{a}-\frac{1}{a+b}\), ta có:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

Câu 2 tương tự nhưng phải phân

\(\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+...+\frac{3}{864}=\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}\)

Cái này áp dụng công thức \(\frac{a}{b\left[b+c\right]}=\frac{a}{c}\left[\frac{1}{b}-\frac{1}{b+c}\right]\), ta có:

\(\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{27}-\frac{1}{32}\right]\)

\(=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{32}\right]=\frac{3}{5}\cdot\frac{15}{32}=\frac{9}{32}\)

Câu 3:

tương tự quy laautj mẫu là 2.5; 5.8 ....

Câu 4: qL mẫu là 1.7; 7.13; ....

Câu 5: \(=\left[1-\frac{1}{2}\right]+\left[1-\frac{1}{6}\right]+...+\left[1-\frac{1}{110}\right]\)

\(=\left[1-\frac{1}{1\cdot2}\right]+\left[1-\frac{1}{2.3}\right]+\left[1-\frac{1}{3\cdot4}\right]+...+\left[1-\frac{1}{10.11}\right]\)

\(=10-\frac{9}{10}=\frac{91}{10}\)

\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)

= \(4.\left(\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)

=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)

= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)

= \(1.\dfrac{8}{27}\)

= \(\dfrac{8}{27}\)

Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)

\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)

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

\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)

\(=2-\frac{1}{5}\)< 2 

Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2

Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

Ta có : \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}=\frac{1}{2}-\frac{2}{27}=\frac{23}{54}\)

Trả lời:

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

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