Question

Giải các bất phương trình sau:

1) \(\dfrac{\text{x}-1}{x-3}>1\)                                2) \(\sqrt{\text{x}^2+x-12}< 8-x\)

Answer 1

1:

ĐKXĐ: x<>3

 \(\dfrac{x-1}{x-3}>1\)

=>\(\dfrac{x-1-\left(x-3\right)}{x-3}>0\)

=>\(\dfrac{x-1-x+3}{x-3}>0\)

=>\(\dfrac{2}{x-3}>0\)

=>x-3>0

=>x>3

2: ĐKXĐ: \(\left[{}\begin{matrix}x>=3\\x< =-4\end{matrix}\right.\)

\(\sqrt{x^2+x-12}< 8-x\)

=>\(\left\{{}\begin{matrix}8-x>=0\\x^2+x-12< \left(8-x\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =8\\x^2+x-12-x^2+16x-64< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =8\\17x-76< 0\end{matrix}\right.\)

=>\(x< \dfrac{76}{17}\)

Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}3< =x< \dfrac{76}{17}\\x< =-4\end{matrix}\right.\)