\(\dfrac{0.\left(3\right)+0.\left(384615\right)+\dfrac{3}{13}x}{0.0\left(3\right)+13}=\dfrac{50}{85}\)

\(\Leftrightarrow\dfrac{\dfrac{28}{39}+\dfrac{3}{13}x}{\dfrac{391}{30}}=\dfrac{10}{17}\)

\(\Leftrightarrow x\cdot\dfrac{3}{13}+\dfrac{28}{39}=\dfrac{23}{3}\)

\(\Leftrightarrow x\cdot\dfrac{3}{13}=\dfrac{271}{39}\)

\(\Leftrightarrow x=\dfrac{271}{9}\)

0,(3) = 1/3

0,0(3) = 1/30

0,(384615) = 5/13 

ngô minh hoàng

\(\Leftrightarrow\dfrac{\dfrac{3}{10}+\dfrac{5}{13}+\dfrac{3}{13}x}{\dfrac{1}{30}+13}=\dfrac{50}{85}\)

\(\Leftrightarrow x\cdot\dfrac{3}{13}+\dfrac{89}{130}=\dfrac{23}{3}\)

=>3/13x=2723/390

hay x=2723/90

Đề bài là gì vậy bạn ??? Tính hay tìm x ?

\(\frac{0,\left(3\right)+0,\left(384615\right)+\frac{3}{13}x}{0,0\left(3\right)+13}\)

\(=\frac{\frac{1}{3}+\frac{5}{13}+\frac{3}{13}x}{\frac{1}{30}+13}=\frac{\frac{1}{3}+\frac{5+3x}{13}}{\frac{391}{30}}=\frac{\frac{13+3\left(5+3x\right)}{39}}{\frac{391}{30}}\)

\(=\frac{\frac{13+15+9x}{39}}{\frac{391}{90}}=\frac{\frac{28+9x}{39}}{\frac{391}{90}}=\frac{28+9x}{39}\cdot\frac{90}{391}\)

P/S : Sai đề trầm trọng

Tìm x nha

P/S : Tìm x vậy kết quả nó bao nhiêu mới tìm được chớ ?

Đề còn thiếu không vậy?

\(a.\)

\(-18-x=-8+\left(13\right)\)

\(\Rightarrow-18-x=5\)

\(\Rightarrow x=-18-5=-23\)

\(b.\)

\(x-43=\left(57-x\right)-50\)

\(\Rightarrow x-43=57-x-50\)

\(\Rightarrow x+x=43+57-50\)

\(\Rightarrow2x=50\)

\(\Rightarrow x=25\)

\(c.\)

\(-4\left(2x+9\right)-\left(-8x+3\right)-\left(x+13\right)=0\)

\(\Rightarrow-8x-36+8x-3-x-13=0\)

\(\Rightarrow-x-52=0\)

\(\Rightarrow-x=52\)

\(\Rightarrow x=-52\)