Bài 1 : 

Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\) ta có : 

\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(A=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

Chúc bạn học tốt ~ 

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(S=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(S=\frac{1}{3}-\frac{1}{8}\)

\(S=\frac{8}{24}-\frac{3}{24}\)

\(S=\frac{5}{24}\)

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(S=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(S=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

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

\(=\frac{8}{24}-\frac{3}{24}\)

\(=\frac{5}{24}\)

S= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

S= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  1/5 -1/6 +1/6 - 1/7 + 1/7 - 1/8

S= 1/2 - 1/ 8 

S= 3/8

S= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8

  = 1/2 - 1/3 + 1/3 - ...+ 1/7 - 1/8

  = 1/2 - 1/8

  = 3/8

S=(1/2.3) +(1/3.4) + (1/4.5) +...+1/7.8

S=1/2-1/3+1/3-1/4 +1/4-1/5+...+1/7-1/8

S=1/2-1/8

S=3/8

1)

A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

   = \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)

   = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

   = \(\frac{1}{5}-\frac{1}{12}\)

   = \(\frac{7}{60}\)

B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

   = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

   = \(\frac{3.4.5.....100}{2.3.4....99}\)

   = \(\frac{100}{2}=50\)

C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)

   = \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)

   = \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)

   = \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)

   = \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

~ Hok tốt ~

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

....

\(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)

\(S=\dfrac{1}{2}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)

\(S=\dfrac{67}{110}\)

bn có nhầm \(\dfrac{1}{2}\) với \(\dfrac{1}{20}koz\)

\(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)

\(S=\dfrac{1}{2}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)

\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)

\(S=\dfrac{7}{10}-\dfrac{1}{11}\)

\(S=\dfrac{67}{110}\)

Mk trình bày lại, cái ở trên bị lỗi nha

tổng của 3 số là:

(50+42+56) : 2 = 74

số thứ nhất là:

74 - 42 = 32

số thứ 2 là:

50 - 32 = 18

số thứ 3 là:

42 - 18 = 24

đ\s.......

khó quá anh ơi em mới lớp 8

Gọi x,y,z là 3 số cần tìm. Ta có hệ phương trình:

\(\hept{\begin{cases}x+y=50\\y+z=42\\z+x=56\end{cases}}\)

Giải hệ trên, ta được:

x=32

y=18

z=24

o tổng $S=1+3^1+3^2+3^3+...........+3^{30}.$

S là số chính phương hay không ?

Bạn vào câu hỏi tương tự nha

A=1/6+1/12+1/20+1/30+1/42+1/56+1/72

A=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9

A=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

A=1/2-1/9

Câu B tương tự nha bạn :333