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

Cho mình sửa xíu x1/5 sửa thành x+1/5

x+1/5-1/45 = -37/45

x+8/45        = -37/45

x               = -37/45-8/45

x                = -1

\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)

\(x+4\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{4}{13.17}+...+\frac{1}{41.45}\right)=\frac{-37}{45}\)

\(x+4.\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(x+1\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(x+\frac{8}{45}=\frac{-37}{45}\)

\(x=\frac{-37}{45}-\frac{8}{45}\)

\(x=\frac{-45}{45}=-1\)

LÀ 4,23 

x có  là 4,23

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

x+ 1/5.9 + ... + 1/41+ 1/45 = -37/45

x  + 1/5 - 1/9 + 1/9 -  ......+ 1/41 - 1/45 = -37/45

x+ 1/5 - 1/45 = -37 45

x+ 8/45 =-37/45

x            = -37 /45 - 8/45

x                = -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:

\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=-\frac{37}{45}\)

\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(x+\frac{8}{45}=-\frac{37}{45}\)

\(x=-\frac{37}{45}-\frac{8}{45}\)

\(x=\frac{-45}{45}=-1\)

\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)

\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)

\(\Leftrightarrow x=-\frac{43}{45}\)

Ta có:

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)

\(=\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)

\(=\frac{7}{x}+\frac{1}{5}-\frac{1}{45}\)

\(=\frac{7}{x}+\frac{9}{45}-\frac{1}{45}=\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\)

\(\Rightarrow x=15\)

Vậy x=15