Dat A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\)

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\)

= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-....+\dfrac{1}{13}-\dfrac{1}{15}\)

= 1-\(\dfrac{1}{15}=\dfrac{14}{15}\)

=> A=\(\dfrac{7}{15}\)

Ta co : \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

=> \(\dfrac{7}{15}x-\dfrac{7}{15}+\dfrac{7}{15}=\dfrac{3}{5}x\)

=> \(\dfrac{7}{15}x-\dfrac{3}{5}x=0\)

=> x\(\left(\dfrac{7}{15}-\dfrac{3}{5}\right)=0\)

=> x\(\left(-\dfrac{2}{15}\right)=0\)

=> x=0

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=> \(\dfrac{7}{15}\left(x-1\right)=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{7}{15}x-\dfrac{7}{15}=\dfrac{3}{5}x-\dfrac{7}{15}\)

<=>\(\dfrac{7}{15}x-\dfrac{3}{5}x=\dfrac{-7}{15}+\dfrac{7}{15}\)

<=> \(\dfrac{-2}{15}x=0\)

<=> \(x=0\)

Vậy: \(s=\left\{0\right\}.\)

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\left(1-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\frac{14}{15}\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\frac{14}{15}x-\frac{14}{15}=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow-\frac{4}{15}x=\frac{28}{15}\)

\(\Leftrightarrow x=7\)

xong r. k cần nữa nhé

\(x=\frac{7}{5}\)

nhớ tick nha pạn

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)

\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)

Tập nghiệm: \(S=\left\{0\right\}\)

a) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}-\frac{3}{5}x+\frac{7}{15}=0\)

\(\Leftrightarrow\frac{8}{15}x=0\)

\(\Leftrightarrow x=0\)

a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0+3/4.x = 9/10

3/4.x = 9/10

x = 9/10: 3/4

x = 6/5

b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5

x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5

x + 3/2. ( 1-1/15) = 11/5

x + 3/2.14/15 = 11/5

x + 7/5 = 11/5

x = 11/5 - 7/5

x = 4/5

..... là gì?