Question

$a)\dfrac{x-3}{x-2}+$$\dfrac{x-2}{x-4}=-1$

Giải pt

Answer 1

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

\(\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-\dfrac{1\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow x^2-7x+12+x^2-4x+4=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\) (thỏa mãn)