a: ĐKXĐ: \(x\ne2\)
\(\dfrac{1}{x-2}=3\)
=>\(x-2=\dfrac{1}{3}\)
=>\(x=2+\dfrac{1}{3}=\dfrac{7}{3}\left(nhận\right)\)
b: ĐKXĐ: \(x\notin\left\{5;\dfrac{3}{2}\right\}\)
\(\dfrac{3}{5-x}-\dfrac{4}{2x-3}=1\)
=>\(-\dfrac{3}{x-5}-\dfrac{4}{2x-3}=1\)
=>\(\dfrac{3}{x-5}+\dfrac{4}{2x-3}=-1\)
=>\(\dfrac{3\left(2x-3\right)+4\left(x-5\right)}{\left(x-5\right)\left(2x-3\right)}=-1\)
=>\(\left(x-5\right)\left(2x-3\right)=-\left[3\left(2x-3\right)+4\left(x-5\right)\right]\)
=>\(\left(2x-3\right)\left(x-5\right)=-\left[6x-9+4x-20\right]\)
=>\(2x^2-10x-3x+15+\left(10x-29\right)=0\)
=>\(2x^2-3x-14=0\)
=>\(2x^2-7x+4x-14=0\)
=>x(2x-7)+2(2x-7)=0
=>(2x-7)(x+2)=0
=>\(\left[{}\begin{matrix}x=\dfrac{7}{2}\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: \(x\notin\left\{5;2;-2\right\}\)
\(\dfrac{3}{x^2-4}-\dfrac{7}{5-x}=2x\)
=>\(\dfrac{3}{x^2-4}+\dfrac{7}{x-5}=2x\)
=>\(\dfrac{3\left(x-5\right)+7\left(x^2-4\right)}{\left(x-5\right)\left(x^2-4\right)}=2x\)
=>\(\dfrac{3x-15+7x^2-28}{\left(x-5\right)\left(x^2-4\right)}=2x\)
=>\(2x\left(x^3-4x-5x^2+20\right)=7x^2+3x-43\)
=>\(2x^4-10x^3-8x^2+40x-7x^2-3x+43=0\)
=>\(2x^4-10x^3-15x^2+37x+43=0\)
=>\(x\in\left\{-1,62;-1,12;2,11;5,63\right\}\)
a, đk x khác 2
\(\dfrac{1}{x-2}=3\Rightarrow1=3x-6\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\left(tm\right)\)
b, đk x khác 5 ; 3/2
\(\dfrac{3}{5-x}-\dfrac{4}{2x-3}=1\Rightarrow6x-9-4\left(5-x\right)=\left(5-x\right)\left(2x-3\right)\)
\(\Leftrightarrow10x-29=-2x^2+13x-15\Leftrightarrow-2x^2+3x+14=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-2\end{matrix}\right.\left(tm\right)\)
c, đk x khác -2 ; 2; 5
\(\dfrac{3}{x^2-4}-\dfrac{7}{5-x}=2x\Rightarrow15-3x-7\left(x^2-4\right)=2x\left(x^2-4\right)\left(5-x\right)\)
\(\Leftrightarrow2x\left(5x^2-x^3-20+4x\right)=15-3x-7x^2+28\)
\(\Leftrightarrow10x^3-2x^4-40x+8x^2=15-3x-7x^2+28\)
\(\Leftrightarrow10x^3-2x^4+15x^2-37x-43=0\)
