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

\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)

\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)

\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)

\(=\dfrac{-5+x}{5x}\)

\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)

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

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

\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)

\(=\dfrac{x^2+x-6}{x^2+x-6}\)

\(=1\)

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

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

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

=> 15x²-102x-5x+34= 1000-120x-125x+15x²

=> 15x²- 107x+34 = 1000-245x+15x²

=> 107x+34 =1000-245x

=> 138x=966

=> x=7

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

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

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

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

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

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

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

\(\Rightarrow138x=966\)

\(\Rightarrow x=7\)

Mình chắc chắn đáp án này đúng.

1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)

Suy ra: \(5x^2+3x-9=5x^2-5x\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(tm\right)\)

2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(6x=3x-15\)

\(\Leftrightarrow3x=-15\)

hay \(x=-5\left(loại\right)\)

2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)

Vậy pt vô nghiệm.

3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)

\(\Rightarrow -3(x+4)-(3-5x)=x-4\)

\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)