Bạn xem lại đề nha

    \(\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)

\(\#PeaGea\)

1)
     \(7,5-\left(\dfrac{1}{3}x+\dfrac{1}{6}\right)=\dfrac{1}{3}\)
\(\Leftrightarrow7,5-\dfrac{1}{3}x-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Leftrightarrow7,5-\dfrac{1}{3}x=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{3}x=7,5-\dfrac{1}{2}=7\)
\(\Leftrightarrow x=7:\dfrac{1}{3}=21\)
    Vậy \(x=21\)


2)
     \(4,5-\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)=\dfrac{1}{6}\)
\(\Leftrightarrow4,5-\dfrac{1}{2}x-\dfrac{1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow4,5-\dfrac{1}{2}x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=4,5-\dfrac{1}{2}=4\)
\(\Leftrightarrow x=4:\dfrac{1}{2}=8\)
    Vậy \(x=8\)

\(\#PeaGea\)

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
     \(=\left(1+\dfrac{1}{5}\right)+\left(\dfrac{3}{5\cdot7}+...+\dfrac{3}{97\cdot99}+\dfrac{3}{99\cdot101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\cdot\dfrac{96}{505}\)
     \(=\dfrac{6}{5}+\dfrac{64}{505}\)
     \(=\dfrac{134}{101}\)

\(\#PeaGea\)

Có ảnh của nút lệnh không bạn?

Bạn xem lại đề nhé bạn

Bài 5:
a)
\(S=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+\dfrac{1}{17\cdot21}+\dfrac{1}{21\cdot25}\)
   \(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}\right)\)
   \(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}\right)\)
   \(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)\)
   \(=\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)
   Vậy \(S=\dfrac{1}{25}\)

b)
    Để \(\dfrac{n-5}{n-3}\) có giá trị là số nguyên thì \(n-5⋮n-3\)
    Ta có:
     \(n-5⋮n-3\)
\(\Leftrightarrow n-3-2⋮n-3\)
\(\Leftrightarrow-2⋮n-3\)
\(\Rightarrow n-3\in\text{Ư}\left(-2\right)=\left\{\pm1;\pm2\right\}\)
    Ta có bảng:
    

Vậy \(n\in\left\{4;2;5;1\right\}\)

\(\#PeaGea\)