a, đk x khác 1
\(\Leftrightarrow x^2+x+1-2x^2+5=4x-4\Leftrightarrow-x^2+x+6=4x-4\)
\(\Leftrightarrow-x^2-3x+10=0\Leftrightarrow x=2;x=-5\)(tm)
b, đk x khác 0 ; -5 ; 5
\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x^2+25x\)
\(\Leftrightarrow x^2+30x+25=x^2+25x\Leftrightarrow5x+25=0\Leftrightarrow x=-5\)(ktm)
Vậy pt vô nghiệm
a: \(\Leftrightarrow x^2+x+1-2x+5=4x-4\)
\(\Leftrightarrow x^2-x+6-4x+4=0\)
\(\Leftrightarrow x^2-5x+10=0\)(vô lý)
b: \(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
=>5x=25
hay x=5(loại)
