Ta có : 

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)

\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)

\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)

\(=>35x^2+17x+2=35x^2+44x-7\)

\(=>17x+2=44x-7\)

\(=>44x-17x=2+7\)

\(=>27x=9\)

\(=>x=\frac{9}{27}=\frac{1}{3}\)

A

\(\frac{1}{3}\)

có 1 giá trị chứ mấy hả bạn

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{7x+2-7x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)

2(7x+2) =5x+7

14x -5x = 7 -4

9x =3

=> x =1/3

Có 1 giá trị của x nhé.

\(\frac{1}{3}\)

\(x=\frac{1}{3}\)

\(x\ne-\frac{7}{5};x\ne-\frac{1}{5}\)

Đề \(\Leftrightarrow\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow35x^2+7x+10x+2=35x^2+49x-5x-7\)

\(\Leftrightarrow17x-44x=-2-7\)

\(\Rightarrow-27x=-9\Rightarrow x=\frac{1}{3}\) (thỏa)

Vậy x = 1/3

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{7x+2-\left(7x-1\right)}{5x+7-\left(5x+1\right)}=\frac{7x+2-7x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)

tu \(\frac{7x+2}{5x+7}=\frac{1}{2}=>\left(7x+2\right).2=5x+7=>14x+4=5x+7=>14x-5x=7-4=>9x=3=>x=\frac{3}{9}=\frac{1}{3}\)

vay x=1/3

tick nha