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

\(\#PeaGea\)

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 \(x=8\)

\(\#PeaGea\)

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

\(\#PeaGea\)

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