\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)

= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)

Ta có :

  \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}.\dfrac{202}{203}\)

\(=\dfrac{505}{203}\)

\(\dfrac{2x-1}{12}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)3=12\cdot5\)
\(\Rightarrow6x-3=60\)
\(\Rightarrow6x=63\)
\(\Rightarrow x=\dfrac{21}{2}\)
\(\dfrac{2x-3}{15}=\dfrac{3}{5}\)
\(\Rightarrow\left(2x-3\right)5=15\cdot3\)
\(\Rightarrow10x-15=45\)
\(\Rightarrow10x=60\)
\(\Rightarrow x=6\)
#Đạt Đang Bận Thở

\(\dfrac{2x-1}{12}=\dfrac{20}{12}\)

\(\Leftrightarrow2x-1=20\)

\(\Leftrightarrow2x=21\)

\(\Leftrightarrow x=\dfrac{21}{2}\)

b. \(\dfrac{2x-3}{15}=\dfrac{9}{15}\)

\(\Leftrightarrow2x-3=9\)

\(\Leftrightarrow x=6\)

\(\dfrac{x}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{y+3}\)  Đk (\(y\ne-3\))⇒ \(\dfrac{2x+3}{6}\) = \(\dfrac{1}{y+3}\) ⇒ (2\(x\)+3)(y+3) = 6

Ư(6) = { -6; -3; -2; -1; 1; 2; 3; 6}

Lập bảng ta có:

Từ bảng trên ta có các cặp \(x\), y nguyên thỏa mãn đề bài là:

(\(x\), y) = ( -3; -5); ( -2; -9); ( -1; 3); (0; -1); 

\(\dfrac{x-7}{3}+\dfrac{x-6}{4}=2\)

\(\Leftrightarrow\dfrac{4\left(x-7\right)}{12}+\dfrac{3\left(x-6\right)}{12}=\dfrac{2.12}{12}\)

\(\Leftrightarrow\dfrac{4x-28+3x-18}{12}=\dfrac{24}{12}\)

\(\Leftrightarrow\dfrac{7x-46}{12}=\dfrac{24}{12}\)

\(\Leftrightarrow7x-46=24\)

\(\Leftrightarrow7x=70\)

\(\Leftrightarrow x=10\)

\(\dfrac{4\left(x-7\right)}{12}+\dfrac{3\left(x-6\right)}{12}=\dfrac{24}{12}\)

\(4x-28+3x-18=24\)

\(7x-46=24\)

\(7x=70\)

\(x=10\)

A = \(\dfrac{3}{3\times7}\)+ \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\)+...+\(\dfrac{3}{107\times111}\)

A = \(\dfrac{3}{4}\) \(\times\)( \(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+ \(\dfrac{4}{11\times15}\)+...+\(\dfrac{4}{107\times111}\))

A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{107}\)- \(\dfrac{1}{111}\))

A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{111}\))

A = \(\dfrac{9}{37}\) > \(\dfrac{9}{45}\) = \(\dfrac{1}{5}\) 

Vậy  \(\dfrac{3}{3\times7}\) + \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\) + ...+ \(\dfrac{3}{107\times111}\) > \(\dfrac{1}{5}\) ( đpcm)

Bạn ơi thế này thì đúng hơn chứ:

\(\dfrac{3}{3.7}+\dfrac{3}{7.11}+\dfrac{3}{11.15}+...+\dfrac{3}{107.111}>\dfrac{1}{5}\)

\(\dfrac{11}{5}-x=\dfrac{3}{5}:\dfrac{5}{3}\)

\(\dfrac{11}{5}-x=\dfrac{9}{25}\)

\(x=\dfrac{11}{5}-\dfrac{9}{25}=\dfrac{46}{25}\)

\(\dfrac{5}{3}\) \(\times\)[ 2\(\dfrac{1}{5}\) - \(x\)]\(^2\) = \(\dfrac{3}{5}\)

       [\(\dfrac{11}{5}\) - \(x\)]²  = \(\dfrac{3}{5}\) : \(\dfrac{5}{3}\)

      [ \(\dfrac{11}{5}\) - \(x\)]² =   \(\dfrac{3}{5}\) \(\times\) \(\dfrac{3}{5}\)

     [ \(\dfrac{11}{5}\) - \(x\)]² = \(\dfrac{9}{25}\)

      \(\left[{}\begin{matrix}\dfrac{11}{5}-x=-\dfrac{3}{5}\\\dfrac{11}{5}-x=\dfrac{3}{5}\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=\dfrac{11}{5}+\dfrac{3}{5}\\x=\dfrac{11}{5}-\dfrac{3}{5}\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=\dfrac{8}{5}\end{matrix}\right.\)

\(x\in\) { \(\dfrac{8}{5};\dfrac{14}{5}\)}

A =\(\dfrac{4^2}{3\times5}\) \(\times\)\(\dfrac{5^2}{4\times6}\) \(\times\) \(\dfrac{6^2}{5\times7}\) \(\times\) \(\dfrac{7^2}{6\times8}\)

A = \(\dfrac{4\times4\times5^2\times6^2\times7\times7}{3\times4\times5^2\times6^2\times7\times8}\)

A =  \(\dfrac{4}{3}\) \(\times\) \(\dfrac{7}{8}\)

A = \(\dfrac{7}{6}\) 

Ta có: 

\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\)        

\(\left[{}\begin{matrix}x-1=0\\y-4=0\\z-9=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0+1\\y=9+4\\z=0+9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\y=4\\z=9\end{matrix}\right.\)