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

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

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

=>(7x+2)*(5x+1)=(5x+7)(7x-1)(t/c 2 p/s bằng nhau)

7x(5x+1)+2(5x+1)=5x(7x-1)+7(7x-1)

35x²+7x+10x+2=35x²-5x+49x-7

17x+2=44x-7

17x-44x=-7-2

-27x=-9

x=-9/-27=1/3

Vậy x=1/3

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

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

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.

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

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

\(2\left(7x+2\right)=5x+7\)

\(14x-5x=7-4\)

\(9x=3\)

\(\Rightarrow x=3\div9\)

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

nhân chéo ta được:

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

\(35x^2+7x+10x+2=35x^2-5x+49x-7\) ( nhân phân phối)

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

\(-27x=-9\)

\(x=\frac{-9}{-27}\)

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

=>x=1/3