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

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

\(\Rightarrow3x\left(5x-34\right)-1\left(5x-34\right)=40\left(25-3x\right)-5x\left(25-3x\right)\)

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

\(\Rightarrow15x^2-107x+34=1000-245x+15x^2\)

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

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

\(\Rightarrow138x=966\)

\(\Rightarrow x=7\)

Mình chắc chắn đáp án này đúng.

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

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

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

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

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)

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=\(\pm\sqrt{7}\)

=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x²+4x+30x+6=20x²+25x+8x+10

=>20x²-20x²+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

B)=>(3x-1).(5x-34)=(40-5x).(25-3x)

=>15x²-102x-5x+34=1000-120x-125x+15x²

=>15x²-107x+34=1000-245x+15x²

=>15x²-15x²-107x+245x=1000-34

=>0-107x+245x=966

=>138x=966

=>x=7

A,=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x2+4x+30x+6=20x2+25x+8x+10

=>20x2-20x2+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

\(\dfrac{3x-1}{40-5x}=\dfrac{25-3x}{5x-34}\\ \Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\\ \Leftrightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\\ \Leftrightarrow15x^2-15x^2-102x-5x+125x+120x=1000-34\\ \Leftrightarrow138x=966\\ \Leftrightarrow x=\dfrac{966}{138}\\ \Leftrightarrow x=7\)

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

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

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

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

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

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

\(\Rightarrow0=966-148x\)

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

áp dụng tính chất của dãy tỉ số bằng nhau ta có;

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

suy ra:

\(\frac{3x-1}{40x-5}=4\Rightarrow3x-1=4.\left(40-5x\right)\)

3x-1=160-20x

3x+20x=160+1

23x=161

x=161:23

x=7

vậy 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!