S=1/3.7+1/7.11+...+1/19.23 (1)

Nhân cả 2 vế của đẳng thức (1) với 4 ta được:

4S=4/3.7+4/7.11+...+4/19.23

4S=1/3.7+1/7.11+...+1/19.23

4S=1/3-1/7+1/7-1/11+..+1/19-1/23

4S=1/3-1/23

4S=20/69

S  =20/69:4

S  =5/69

Mọi người ủng hộ mik nha

\(S=\frac{1.4}{3.7.4}+\frac{1.4}{7.11.4}+......+\frac{1.4}{19.23.4}\)

     \(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+......+\frac{4}{19.23}\right)\)

     \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{19}-\frac{1}{20}\right)\)

      \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{20}\right)\)

     \(=\frac{1}{4}.\frac{17}{60}=\frac{17}{240}\)

\(S=\frac{1.4}{3.7.4}+\frac{1.4}{7.11.4}+......+\frac{1.4}{19.23.4}\)

    \(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+......+\frac{4}{19.23}\right)\)

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

   \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{23}\right)\)

   \(=\frac{1}{4}.\frac{20}{69}=\frac{5}{69}\)

\(=\dfrac{1}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{51.55}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{55}\right)=\dfrac{1}{4}\times\dfrac{52}{165}=\dfrac{13}{165}\)

Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)

=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

k nha bạn

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:

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

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

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

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

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

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

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

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

Học tốt 

Chắc là đề thiếu: \(y=\frac{1}{2}-\frac{1}{3\cdot7}-\frac{1}{7\cdot11}-\frac{1}{11\cdot15}-\frac{1}{15\cdot19}-\frac{1}{19\cdot23}-\frac{1}{23\cdot27}\)

\(y=\frac{1}{2}-\left(\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+...+\frac{1}{23\cdot27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{23\cdot27}\right)\)

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

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

x = \(\frac{163}{528}\)

cho 1 đ-ú-n-g nha bạn

Giải:

A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)

A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)

A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)

A=1/2.(1/3-1/n+4)

A=1/6-1/2.(n+4)

⇒A<1/6

Chúc bạn học tốt!

Ta có : \(A=\dfrac{2}{3.7}+\dfrac{2}{7.11}+...+\dfrac{2}{n\left(n+4\right)}\)

\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)

\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)\(=\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{2}{11}+...+\dfrac{2}{n}-\dfrac{2}{n+4}=\dfrac{2}{3}-\dfrac{2}{n+4}\)

\(\Rightarrow A=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}\)

- Xét hiệu \(A-\dfrac{1}{6}=-\dfrac{1}{2\left(n+4\right)}< 0\)

Vậy A < 1/6

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

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

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

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

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

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

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

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