Question

1) Giải hệ phương trìnhb)\(\left\{{}\begin{matrix}\dfrac{x-1}{x 1} \dfrac{y-2}{y 1}=2\\\dfrac{2}{x 1} \dfrac{2}{y 1}=1\end{matrix}\right.\)

Answer 1

a) ĐKXĐ: \(x\notin\left\{4;-1\right\}\)

Ta có: \(\dfrac{2x}{x^2-3x-4}-\dfrac{x}{x+1}=\dfrac{3}{x-4}\)

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

Suy ra: \(2x-x^2+4x=3x+3\)

\(\Leftrightarrow-x^2+6x-3x-3=0\)

\(\Leftrightarrow-x^2+3x-3=0\)

\(\Delta=3^2-4\cdot\left(-1\right)\cdot\left(-3\right)=9-12=-3\)

Vì \(\Delta< 0\) nên phương trình vô nghiệmVậy: \(S=\varnothing\)