a)\(\frac{2x-5}{x+5}\)=3 ĐKXĐ: x khác -5
=> 2x-5=3(x+5)
<=>2x-5=3x+15
<=>-x=20
<=>x =-20
b)\(\frac{x2-6}{x}\)=x+\(\frac{3}{2}\)ĐKXĐ\(x\ne0\)
=>2(x2-6)=2x2+3x
<=>2x2-12=2x2+3x
<=>-3x=12
<=>x=-4
c)\(\frac{x2+2x-3x-6}{x-3}\)=0
chia cả 2 vế cho x-3 ta được pt mới
x2+2x-3x-6=0
<=>x2-x-6=0
<=>x2-3x+2x-6=0
<=>(x2-3x)+(2x-6)=0
<=>x(x-3)+2(x-3)=0
<=>(x-3)(x+2)=0
<=>\(\begin{cases}x=3\\x=-2\end{cases}\)
e)\(\frac{2x-1}{x-1}+1=\frac{1}{x-1}\)\(ĐKXĐ:x\ne1\)
<=>2x-1+x-1=1
<=>3x=3
<=>x=1
d)\(\frac{5}{3x+2}=2x-1\)ĐKXĐ:\(x\ne-\frac{2}{3}\)
<=>5=(2x-1)(3x+2)
<=>5=6x2+4x-3x-2
<=>-(6x2-4x+3x-7)=0
<=>6x2-x-7=0
<=>6x2+6x-7x-7=0
<=>6x(x+1)-7(x+1)=0
<=>(x+1)(6x-7)=0
<=>\(\begin{cases}x=-1\\x=\frac{7}{6}\end{cases}\)
\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)\(ĐKXĐ:x\ne0;x\ne-1\)
=>x(x+3)+(x-2)(x+1)=2x(x+1)
<=>x2+3x+x2-x-2=2x2+2x
<=>0x=2( vô lí)
=> pt vô nghiệm