Ta có;

\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}=\frac{3x-1+25-3x}{40-5x+5x-34}=\frac{24}{6}=4\)

=> 25 - 3x = 4(5x - 34)

=> 25 - 3x = 20x - 136

=> 25 + 136 = 20x + 3x

=> 161 = 23x

=> x = 7

  với đề là (3x - 1)/40 - 5x = (25 -3x)/5x-3 thì nhân chéo lên mà giải thui mà: 
(3x -1).(5x -34 )=(40 - 5x).(25 - 3x) 
=>x=7

\(x\ne8;x\ne\frac{34}{5}\)

\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\)

\(\Rightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\)

\(\Rightarrow-102x-5x+125x+120x=1000-34\)

\(\Rightarrow138x=966\Rightarrow x=7\)

Vậy x = 7

x = 7

tick mình nha!

mik nhầm nha, phải là:

138x=966

=> x=7

Ta có : 

\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(=>\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)

\(=>15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(=>15x^2-107x+34=1000-245x+15x^2\)

\(=>34-107x=1000-245x\)

\(=>1000-245x+107x=34\)

\(=>1000-138x=34\)

\(=>138x=1000-34=966\)

\(=>x=\frac{966}{138}=7\)

theo đề bài ta có: \(\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)

=> \(15x^2-107x+34=1000-245x+15x^2\)

<=> 138*x^2=966

<=>x^2=7

<=>x=\(\pm\sqrt{7}\)

\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)

\(\Rightarrow3x\left(5x-34\right)-\left(5x-34\right)=40\left(25-3x\right)-5x\left(25-3x\right)\)

\(\Rightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Rightarrow15x^2-97x+34=1000-245x+15x^2\)

\(\Rightarrow15x^2=1000-34-245x+97x+15^2\)
\(\Rightarrow15x^2=966-148x+15^x\)

\(\Rightarrow0=966-148x\)

\(\Rightarrow x=\frac{996}{148}=\frac{249}{37}\)

\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)