\(a,2\left(x-3\right)-5\left(2x-4\right)=0\)
=> \(2x-6-10x-20=0\)
=> \(\left(2x-10x\right)-\left(6+20\right)=0\)
=> \(-8x-26=0\)
=> \(-8x=26\)
=> \(x=26:-8=-\frac{13}{4}\)
Vậy \(x\in\left\{-\frac{13}{4}\right\}\)
\(b,3+\frac{1}{x-8}=0\)
=> \(\frac{1}{x-8}=0-3=-3\)
=> \(x-8=-\frac{1}{3}\)
=> \(x=-\frac{1}{3}+8=\frac{23}{3}\)
Vậy \(x\in\left\{\frac{23}{3}\right\}\)
\(c,\frac{8}{3}-\frac{2x+3}{5}=\frac{-7}{3}\)
=> \(15.\frac{8}{3}-15.\frac{2x+3}{5}=15.\frac{-7}{3}\)
Chiệt tiêu
=> \(5.8-3\left(2x+3\right)=5.\left(-7\right)\)
=> \(40-\left(6x+9\right)=-35\)
=> \(40-6x-9=-35\)
=>\(31=6x=-35\)
=> \(6x=41-\left(-35\right)=66\)
=> \(x=66:6=11\)
Vậy \(x\in\left\{11\right\}\)
\(d,\frac{1}{9}=\frac{5}{3x-5}=0\)
=> \(\frac{1}{9}=0\left(sai\right)\)
=> \(x\in\varnothing\)
