\(c,ĐK:x\ne\dfrac{5}{4}\\ \dfrac{2x+3}{4x-5}=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\\ d,ĐK:x\ne1\\ \dfrac{x^2-1}{x^2-2x+1}=0\\ \Leftrightarrow x^2-1=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=-1\)
c, ĐK:x khác 5/4
Ta có 2x+3/4x-5=0=> 2x +3 =0=> x=-3/5
d, ta có x^2-1/x^2-2x+1=x+1/x-1 (ĐK : x khác 1)
Để x^2-1/x^2-2x+1=0 =>x+1/x-1=0
=>x=-1