a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

b) 

 \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

        \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)

         \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

          \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Vậy \(x=-329\)

b) \(\left|5x-3\right|-x=7\)

\(\Rightarrow\left|5x-3\right|=7+x\)

\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)

Vậy .................... 

Bạn ơi !!! ý A tham khảo tại link này nè :

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a) Ta thấy:
\(\left(x+4\right)\left(x-4\right)=x\left(x-\frac{2}{3}\right)\)
\(\Rightarrow\left(x^2-4x\right)+\left(4x-16\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow\left(x^2-16\right)-\left(4x-4x\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16-0=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16=x^2-\frac{2}{3}x\)
\(\Rightarrow16=\frac{2}{3}x\)    ( do có cùng hiệu và cùng số bị trừ )
\(\Rightarrow x=16:\frac{2}{3}\)
\(\Rightarrow x=24\)
Vậy x = 24

b.) x^3-x^2-2x=0

    x(x^2-x-2)=0

   x(x^2-2x+x-2)=0

   x(x(x-2)+x-2)=0

  x(x-2)(x+1)=0

suy ra x=0 hoặc x-2=0 hoặc x+1=0 

    vậy x=0 hoặc x=2 hoặc x=-1 

hình như câu c đề phải là (x+4)/120 thì phải đó bạn 

c.)(x+4)/120+(x+8)/116=(x+5)/119+(x+7)/117

   (x+4)/120+(x+8)/116-(x+5)/119-(x+7)/117=0

   (x+4)/120+1+(x+8)/116+1-(x+5)/119-1-(x+7)/117-1=0

   (x+4)/120+1+(x+8)/116+1-((x+5)/119+1)-((x+7)/117+1)=0

   (x+124)/120+(x+124)/116-(x+124)/119-(x+124)/117=0

(x+124)(1/120+1/116-1/119-1/117)=0

suy ra x+124=0

 x=-124

Ta có: 2x + 3y + 5z - 119 = 0

=>  2x + 3y + 5z = 119

 \(\frac{x+2}{3}=\frac{y+3}{5}=\frac{z-4}{7}\Leftrightarrow\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}\)

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

\(\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}=\frac{2x+4+3y+9+5z-20}{6+15+35}=\frac{119+4+9-20}{56}=\frac{112}{56}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x+2}{3}=2\\\frac{y+3}{5}=2\\\frac{z-4}{7}=2\end{cases}\Rightarrow}\hept{\begin{cases}x+2=6\\y+3=10\\z-4=14\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=7\\z=18\end{cases}}\)

Vậy...

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

a) 4/3 - x = 3/5 + 1/2

=> 4/3 - x= 0,8

=> x = 4/3 + 0/8 

=> x = 5/8

b) ( 1/2 + 1/3 +1/6)x = 2/9

= 1x = 2/9 

=> x = 2/9

không biết

Bài 1 :

a) \(\frac{x}{7}=\frac{18}{14}\)

=> x.14 = 7.18

x.14 = 126

x = 126:14

x = 9

b) \(\frac{6}{x}=\frac{7}{4}\)

=> \(x=\frac{6.4}{7}=\frac{24}{7}\)

c) Theo mình đề thế này mới đúng \(\frac{5,7}{0,35}=\frac{\left(-x\right)}{0,45}\)

=> 5,7.0,45 = 0,35.(-x)

2,565 = 0,35.(-x)

(-x) = 2,565:0,35

(-x) = 513/70

=> -x = -513/70

x = 513/70

Bài 2 : Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x-y+z}{2-4+6}=\frac{8}{4}=2\)

\(\frac{x}{2}=2\) 

x = 2.2

x = 4

\(\frac{y}{4}=2\)

y = 2.4

y = 8

\(\frac{z}{6}\) = 2

z = 2.6

z = 12

Vậy x=4 ; y=8 và z=12

\(\frac{x}{7}=\frac{18}{14}\Rightarrow x=18\cdot7:17=9\)

a, (x-15):5+22=24

    ( x - 15 ) : 5 = 2

      x-15  = 10

       x = 25