b: ĐKXĐ: \(x\in R\)
\(\dfrac{4x-17}{2x^2+1}=0\)
=>4x-17=0
=>4x=17
=>\(x=\dfrac{17}{4}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{2;1\right\}\)
\(1+\dfrac{2x-5}{x-2}-\dfrac{3x-5}{x-1}=0\)
=>\(\dfrac{\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\)
=>\(\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\)
=>\(x^2-3x+2+2x^2-9x+5-3x^2+11x-10=0\)
=>-x-3=0
=>x=-3(nhận)
f: ĐKXĐ: \(x\notin\left\{2;4\right\}\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
=>\(\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\)
=>\(\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)\)
=>\(x^2-7x+12+x^2-4x+4=-\left(x^2-6x+8\right)\)
=>\(2x^2-11x+16+x^2-6x+8=0\)
=>\(3x^2-17x+24=0\)
=>(x-3)(3x-8)=0
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=\dfrac{8}{3}\left(nhận\right)\end{matrix}\right.\)
