Question

[Tsumi Akochi]

giải bpt:

a) \(\dfrac{2x^2+10}{1-x}\)\(\le\)0

b)\(\dfrac{3x-4}{x+2}\)\(\ge\)4

c)\(\dfrac{1}{x+4}\)\(\le\)\(\dfrac{1}{x-2}\)

Answer 1

[Mỹ Duyên]

Mk thấy mấy cái này dễ mà, toàn trong sách giáo khoa hết á. Bạn cố gắng đọc và lm đi. Sắp lên lớp 9 rồi đó

Answer 2

[Chí Cường]

a)\(\dfrac{2x^2+10}{1-x}\le0\Rightarrow1-x< 0\Leftrightarrow x>1\)

b) \(\dfrac{3x-4}{x+2}\ge4\Leftrightarrow\dfrac{3x-4}{x+2}-\dfrac{4\left(x+2\right)}{x+2}\ge0\Leftrightarrow\dfrac{-x-12}{x+2}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x-12\le0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-12\\x< -2\end{matrix}\right.\Leftrightarrow-12\le x< -2}}\\\left\{{}\begin{matrix}-x-12\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-12\\x>-2\end{matrix}\right.\end{matrix}\right.\)\(S=\left\{x|-12\le x< -2\right\}\)

c) \(\dfrac{1}{x+4}\le\dfrac{1}{x-2}\Leftrightarrow\dfrac{6}{\left(x+4\right)\left(x-2\right)}\le0\Rightarrow\left(x+4\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 2\end{matrix}\right.\Leftrightarrow-4< x< 2}}\\\left\{{}\begin{matrix}x+4< 0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x>2\end{matrix}\right.\end{matrix}\right.\)

\(S=\left\{x|-4< x< 2\right\}\)