- A ở trên giữa các phân số là dấu " + " nha mấy bạn !

c)  \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\) 

\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\) 

\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=2\left(1-\frac{1}{16}\right)\) 

\(=2.\frac{15}{16}\) 

\(=\frac{15}{8}\) 

Vậy A=\(\frac{15}{8}\)

a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

\(=3.\frac{99}{100}=\frac{297}{100}\)

\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)

=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)

=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)

=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)

\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)

\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)

\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)

\(S\times4=\frac{1}{5}-\frac{1}{25}\)

\(S\times4=\frac{4}{25}\)

\(S=\frac{1}{25}\)

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25

4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25

4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)

4A= 1/5- 1/25

4A= 4/25

A= 4/25 :4 

A= 1/25

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)

\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}.\frac{20}{21}=\frac{7.4.5}{4.7.3}\)

\(=\frac{5}{3}\)

~ Rất vui vì giúp đc bn ~

Bài giải

\(\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+\frac{7}{9\cdot13}+\frac{7}{13\cdot17}+\frac{7}{17\cdot21}\)

\(=\frac{7}{4}\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\right)\)

\(=\frac{7}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\left(1-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}\)

\(=\frac{35}{21}\)

\(a,\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow x^2=8^2\)

\(\Leftrightarrow x=\pm8\)

\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Leftrightarrow x=15\)

Vậy x = 15

Bài cuối tương tự

Ta có:

\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)

b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)

\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)

\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)

\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)

Bn Tấn sai rùi

phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)

\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)

\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\) 

\(4A=1-\frac{1}{x+4}\) 

\(4A=\frac{x+4-1}{x+4}\)   

\(A=\frac{x+3}{\text{4(x+4)}}\)

Bạn tự thay rồi tính nhé 

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)

\(4A=1-\frac{1}{x+4}\)

\(A=\left(1-\frac{1}{x+4}\right):4\)

Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)

A = \(\left(1-\frac{1}{16}:4\right)\)

A = \(\frac{15}{16}:4=\frac{15}{64}\)

Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)

A = \(\left(1-\frac{1}{6}\right):4\)

A \(=\frac{5}{6}:4=\frac{5}{24}\)

Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)

A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)

A = \(\frac{23}{29}:4=\frac{23}{116}\)