a)\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
\(\left(x-\dfrac{1}{2}\right).3+\dfrac{1}{5}=\dfrac{68}{7}\)
\(3x-\dfrac{3}{2}+\dfrac{1}{5}=\dfrac{68}{7}\)
\(3x-\dfrac{13}{10}=\dfrac{68}{7}\)
\(3x=\dfrac{771}{70}\)
\(x=\dfrac{257}{70}\)
Vậy \(x=\dfrac{257}{70}\)
b)\(\dfrac{75}{100}x-x=-1\dfrac{3}{4}\)
\(-\dfrac{1}{4}x=-\dfrac{7}{4}\)
\(x=7\)
Vậy \(x=7\)
a) Ta có: \(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=\dfrac{68}{7}-\dfrac{1}{5}=\dfrac{340}{35}-\dfrac{7}{35}=\dfrac{333}{35}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{333}{35}\cdot\dfrac{1}{3}=\dfrac{111}{35}\)
hay \(x=\dfrac{257}{70}\)
Vậy: \(x=\dfrac{257}{70}\)
b) Ta có: \(75\%x-x=-1\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{4}x=-\dfrac{7}{4}\)
hay x=7
Vậy: x=7
Cho xOy và yOz là hai góc kề bù,biết xOy=50 độ.Vẽ tia Ot là phân giác của xOy.Vẽ tia Om nằm giữa hai tia Oy,Oz sao cho tOm=90 độ.
a)Tính yOm
b)Tia Om có là tia phân giác của yOz không?Vì sao?
a) \(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=\dfrac{68}{7}-\dfrac{1}{5}\)
\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=\dfrac{333}{35}\)
\(x-\dfrac{1}{2}=\dfrac{333}{35}.\dfrac{1}{3}\)
\(x-\dfrac{1}{2}=\dfrac{111}{35}\)
\(x=\dfrac{333}{5}+\dfrac{1}{2}\)
\(x=\dfrac{257}{70}\)
b) \(75\%x-x=-1\dfrac{3}{4}\)
\(x.\left(75\%-1\right)=\dfrac{-7}{4}\)
\(x.\dfrac{-1}{4}=\dfrac{-7}{4}\)
\(x=\dfrac{-7}{4}:\dfrac{-1}{4}\)
\(x=7\)
