<=>\(\frac{\left(2x+1\right).\left(x+1\right)}{\left(x-3\right).\left(x+1\right)}\)+\(\frac{\left(3-2x\right).\left(x-3\right)}{\left(x+1\right).\left(x-3\right)}\)=0

<=>\(\frac{2x^2+3x+1}{\left(x-3\right).\left(x+1\right)}\)+\(\frac{3x-9-2x^2+6x}{\left(x+1\right).\left(x-3\right)}\)=0

<=>\(\frac{2x^2+3x+1+3x-9-2x^2+6x}{\left(x-3\right).\left(x+1\right)}\)=0

<=>\(\frac{12x-8}{\left(x-3\right).\left(x+1\right)}\)=0

=> 12x-8=0

=>x=2/3

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x}{6}=\frac{y}{15}\)( 1 )

\(\frac{y}{3}=\frac{5z}{9}\Rightarrow\frac{y}{15}=\frac{z}{9}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{9}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\Rightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{9}=\frac{3x+2y-z}{18+30-9}=\frac{-78}{39}=-2\)

\(\Rightarrow x=-12;y=-30;z=-18\)

\(\frac{x}{2}\)= \(\frac{y}{5}\); \(\frac{y}{3}\)= \(\frac{5z}{9}\)và 3x+2y-z=-78

\(\Rightarrow\)\(\frac{x}{6}\)= \(\frac{y}{15}\); \(\frac{y}{15}\)\(\frac{5z}{45}\) và 3x+2y-z=-78

\(\Rightarrow\)\(\frac{x}{6}\)= \(\frac{y}{15}\)= \(\frac{5z}{45}\) và 3x+2y-z=-78

\(\Rightarrow\)\(\frac{3x}{18}\)= \(\frac{2y}{30}\)= \(\frac{z}{9}\) và 3x+2y-z=-78

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{3x}{18}\)= \(\frac{2y}{30}\)= \(\frac{z}{9}\)= \(\frac{3x+2y-z}{18+30-9}\)= \(\frac{-78}{39}\)= -2

Suy ra:    \(\frac{x}{6}\)= -2 \(\Rightarrow\)x= 6.(-2)=-12

               \(\frac{y}{15}\)= -2 \(\Rightarrow\)y= 15.(-2)=-30

               \(\frac{z}{9}\)= -2 \(\Rightarrow\)z= 9.(-2)=-18

a, (3x+2)/(5x+7)=3/4

=>4(3x+2)=3(5x+7)

=>12x+8=15x+21

=>12x-15x=21-8

=>-3x=13

=>x=-1

Vậy...

b, 2x-(x-1/7)=0

=>2x-x+1/7=0

=>x=-1/7

Vậy....

a,=> (3x+2).4=(5x+7).3

=>12x+8=15x+21

=>12x+8-15x-21=0

=>-3x=13

=.x=\(\frac{-13}{3}\)

b,<=>2x-x+1/7=0

<=>x=-1/7