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Trả lời:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2006}\)
\(\Rightarrow5x+6=2006\)
\(\Rightarrow5x=2000\)
\(\Rightarrow x=400\)
Vậy x = 400
Trả lời:
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)\(\frac{5}{8}\)
Đặt \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\), ta được : \(\frac{x}{2008}-A=\frac{5}{8}\) (*)
\(\Rightarrow A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(\Rightarrow A=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Thay A vào (*) , ta có:
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
Vậy x = 2008
\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
D
Chọn D