Question

giải phương trình :\(\frac{x-5}{x-1}+\frac{2}{x-3}=1\) biết ĐKXĐ x khác 1 và x khác 3

Answer 1

\(PT< =>\frac{\left(x-5\right)\left(x-3\right)+2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=1\)

<=> \(\frac{x^2-8x+15+2x-2}{\left(x-1\right)\left(x-3\right)}=1\)

<=> \(x^2-6x+13=x^2-4x+3\)

<=> \(x^2-6x+13-x^2+4x-3=0\)

<=> \(-2x+10=0\)

<=> x = 5 (TMDK)

Answer 2

ĐKXĐ: x∉{1;3}

Ta có: \(\frac{x-5}{x-1}+\frac{2}{x-3}=1\)

\(\Leftrightarrow\frac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-8x+15+2x-2=x^2-4x+3\)

⇔\(x^2-6x+13-x^2+4x-3=0\)

\(\Leftrightarrow-2x+10=0\)

⇔\(-2x=-10\)

hay x=5(tm)

Vậy: x=5