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\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{5}-\frac{1}{16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2.\frac{3}{16}]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Rightarrow x=15\)
Theo đề bài ta có
x+1/5-1/9+1/9-1/13+.........+1/41-1/45=-37/45
x+(1/5-1/45)=-37/45
x+8/45=-37/45
x=-37/45 - 8/45
x=-45/45
x=-1
Theo bài ra ta có:
x1/5-1/9+1/9-1/13+...+1/41-1/45=-37/45
x+8/45=-37/45
x=-45/45
x=-1/1
x=-1
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+4\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{4}{13.17}+...+\frac{1}{41.45}\right)=\frac{-37}{45}\)
\(x+4.\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+1\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
bài 1
\(2A=\left(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{99\cdot101}\right)\cdot2\)
\(=5\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\right)\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=5\left(1-\frac{1}{101}\right)\)
\(=5\cdot\frac{100}{101}\)
\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)
bài 2:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x+\dfrac{8}{45}=\dfrac{-37}{45}\\ \Leftrightarrow x=-1\)
theo đề bài ta có:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=1\)
đặt A=4/5.9+4/9.13+4/13.17+...+4/41.45
=1/5-1/9+1/9-1/13+1/13-1/17+...+1/41-1/45
=1/5-1/45
=8/45
suy ra x+8/45=-37/45
suy ra x=-1
a, \(x-\frac{8}{9}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{8}{9}\)
\(\Leftrightarrow x=\frac{11}{9}\)
b, \(\frac{-4}{5}-\frac{8}{15}=\frac{-1}{3}-x\)
\(\Leftrightarrow\frac{-4}{3}=\frac{-1}{3}-x\)
\(\Leftrightarrow x=1\)
c, \(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
Đặt \(A=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)
\(A=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(A=\frac{1}{5}-\frac{1}{45}=\frac{8}{45}\)
Thay A vào phép tính
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\)
\(\Rightarrow x=-1\)