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
Trả lời:
h, \(\left(\frac{1}{2}\right)^{2x}+\left(\frac{1}{4}\right)^x=\frac{1}{8}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2x}+\left[\left(\frac{1}{2}\right)^2\right]^x=\frac{1}{8}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2x}+\left(\frac{1}{2}\right)^{2x}=\frac{1}{8}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{2x}.\left(1+1\right)=\frac{1}{8}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2x}.2=\frac{1}{8}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2x}=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{2x}=\left(\frac{1}{2}\right)^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vây x = 2
Trả lời:
i, \(x-\frac{3}{1.5}-\frac{3}{5.9}-\frac{3}{9.13}-...-\frac{3}{17.21}=\frac{2}{7}\)
\(\Rightarrow x-\left(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{17.21}\right)=\frac{2}{7}\)
Đặt B = \(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{17.21}\) , ta được : \(x-B=\frac{2}{7}\) (**)
\(\Rightarrow B=\frac{3.4}{1.5.4}+\frac{3.4}{5.9.4}+\frac{3.4}{9.13.4}+...+\frac{3.4}{17.21.4}\)
\(=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)
\(=\frac{3}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{3}{4}.\left(1-\frac{1}{21}\right)=\frac{3}{4}.\frac{20}{21}=\frac{5}{7}\)
Thay B = 5/7 vào (**) , ta có:
\(x-\frac{5}{7}=\frac{2}{7}\)
\(\Rightarrow x=1\)
Vậy x = 1
