c: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
=>(x-2)(x+7)=(x+4)(x-1)
\(\Leftrightarrow x^2+5x-35=x^2+3x-4\)
=>5x-35=3x-4
=>2x=31
hay x=31/2
h: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x-3\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2-9x+10x-6=15x^2+21x-5x-7\)
=>x-6=16x-7
=>-15x=-1
hay x=1/15