a: Ta có: \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)
=>\(\frac{\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\)
=>(x-2)(x-1)+(2x-5)(x-1)-(3x-5)(x-2)=0
=>\(x^2-3x+2+2x^2-7x+5-\left(3x^2-6x-5x+10\right)=0\)
=>\(3x^2-10x+7-3x^2+11x-10=0\)
=>x-3=0
=>x=3(nhận)
b: \(\begin{cases}3x+y=7\\ x-7y=-13\end{cases}\Rightarrow\begin{cases}3x+y=7\\ 3x-21y=-39\end{cases}\)
=>\(\begin{cases}3x+y-3x+21y=7+39\\ 3x+y=7\end{cases}\Rightarrow\begin{cases}22y=46\\ 3x=7-y\end{cases}\)
=>\(\begin{cases}y=\frac{23}{11}\\ 3x=7-\frac{23}{11}=\frac{77}{11}-\frac{23}{11}=\frac{54}{11}\end{cases}\Rightarrow\begin{cases}y=\frac{23}{11}\\ x=\frac{18}{11}\end{cases}\)
c: \(3-2x-\frac{9x+6}{3}>0\)
=>3(3-2x)-9x-6>0
=>9-6x-9x-6>0
=>-15x+3>0
=>-15x>-3
=>\(x<\frac{3}{15}=\frac15\)
