d,
(=) (2-3x)(x+11+2-5x)=0
(=) (2-3x)(13-4x)=0
(=)(2-3x)=0hoặc (13-4x)=0
giải từng pt ra là đc
\(d,\left(2-3x\right)\left(x+11\right)\left(2-3x\right)\left(2-5x\right)=0.\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}2-3x=0.\\-4x+13=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=\dfrac{13}{4}.\end{matrix}\right.\)
\(e,x^2-3x+2=0.\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0.\\x-1=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(h,\dfrac{1}{x-1}-\dfrac{2x^2-5}{1-x^3}=\dfrac{4}{x^2+x+1}\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{1}{x-1}+\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{4}{x^2+x+1}=0.\\ \Leftrightarrow\dfrac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0.\\ \Leftrightarrow\dfrac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}=0.\\ \Leftrightarrow\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0.\\ \Leftrightarrow\dfrac{3x}{x^2+x+1}=0.\\ \Rightarrow3x=0.\\ \Leftrightarrow x=0\left(TM\right).\)
