ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)
\(\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\dfrac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\dfrac{x^2-7x+12}{x^2-6x+8}+\dfrac{x^2-4x+2}{x^2-6x+8}=\dfrac{-x^2+6x-8}{x^2-6x+8}\)
\(\dfrac{x^2-7x+12+x^2-4x+2+x^2-6x+8}{x^2-6x+8}=0\)
\(\dfrac{3x^2-17x+22}{x^2-6x+8}=0\)
\(\dfrac{\left(3x-11\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)
\(\dfrac{3x-11}{x-4}=0\)
Với x khác 4
=> 3x -11 = 0
=> \(x=\dfrac{11}{3}\)
hiện tại em đang cần gấp lắm ạ