\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?

\(\dfrac{x-1}{8}=\dfrac{4}{x-2}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)}{8.\left(x-2\right)}=\dfrac{4.8}{\left(x-2\right).8}\)

=>khử mẫu

=>(x-1)(x-2)=32

.......

Là \(\dfrac{x+1}{8}=\dfrac{4}{x-2}\) nhoa

\(a)6/4=9/4\) =)??

thiếu dữ liệu

\(\dfrac{x^2}{20}=\dfrac{4}{5}\)

\(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)

\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)

\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)

\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)

\(x=\dfrac{7}{25}+\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\\ \Rightarrow x=\dfrac{2}{25}\\ x=\dfrac{5}{11}+\dfrac{4}{-9}\\ \Rightarrow x=\dfrac{-45}{-99}+\dfrac{44}{-99}\\ \Rightarrow x=\dfrac{45}{99}+\dfrac{44}{99}\\ \Rightarrow x=\dfrac{95}{99}\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-3}{9}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{8}{9}\Rightarrow x\cdot9=-1\cdot8=-8\\ \Rightarrow x\cdot9=-8\\ \Rightarrow x=\dfrac{-8}{9}\)

` x : 1 / 2 + x : 1 / 3 + x : 1 / 4 + x = 2010`

`x . 2 + x . 3 + x . 4 + x = 2010`

`x ( 2 + 3 + 4 + 1 ) = 2010`

`x . 10 = 2010`

`x = 2010 : 10`

`x = 201`

Vậy ` x= 201`

\(=>x=2010:\left(2+3+4\right)=2010:9=\dfrac{670}{3}\)

ta có : `x/2 = y/3 = z/4=> (2x)/4 =(3y)/9 = z/4`

`=> (2x)/4 =(3y)/9 = z/4` và `2x + 3y - z = 27`

Áp dụng t/c dãy tỉ số bằng nhau ta có:

`(2x)/4 =(3y)/9 = z/4 =(2x + 3y - z)/(4+9-4)=27/9=3`

`=>x/2=3=>x=3.2=6`

`=>y/3=3=>x=3.3=9`

`=>z/4=3=>z=3.4=12`

\(\Leftrightarrow xy=63\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)

\(\Leftrightarrow\left(x+2\right)^3=343\)

=>x+2=7

hay x=5