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

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

=>3/13x=2723/390

hay x=2723/90

0,(3) = 1/3

0,0(3) = 1/30

0,(384615) = 5/13 

ngô minh hoàng

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?

\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)