a: \(-\dfrac{1}{2}:\left(2x-1\right)=0,2:\dfrac{-3}{5}\)
=>\(-\dfrac{1}{2}:\left(2x-1\right)=\dfrac{1}{5}\cdot\dfrac{-5}{3}=\dfrac{-1}{3}\)
=>\(2x-1=\dfrac{1}{2}:\dfrac{1}{3}=\dfrac{3}{2}\)
=>\(2x=\dfrac{3}{2}+1=\dfrac{5}{2}\)
=>\(x=\dfrac{5}{4}\)
b: \(\dfrac{x+1}{4}=\dfrac{9}{x+1}\)
=>\(\left(x+1\right)^2=9\cdot4=36\)
=>\(\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
=>Chọn x=-7
c: Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\)
=>x=2k; y=5k
xy=10
=>\(2k\cdot5k=10\)
=>10k2=10
=>k2=1
=>\(\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
TH1: k=1
=>\(\left[{}\begin{matrix}x=2\cdot1=2\\y=5\cdot1=5\end{matrix}\right.\)
TH2: k=-1
=>\(\left[{}\begin{matrix}x=2\cdot\left(-1\right)=-2\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)
Vì x>0 và y>0 nên chọn x=2 và y=5
=>x-y=2-5=-3
d: \(\dfrac{3}{1-2x}=\dfrac{-5}{3x-2}\)
=>\(\dfrac{3}{2x-1}=\dfrac{5}{3x-2}\)
=>3(3x-2)=5(2x-1)
=>9x-6=10x-5
=>-x=1
=>x=-1
