Câu hỏi của Ely Trần

Question

tìm x để các bt sau có giá trị âm:

1)$x^2-\dfrac{2}{5}x$       2)$\dfrac{x^2-1}{x^2}$

Đức Hiếu · Accepted Answer

a, $x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x>\dfrac{2}{5}\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x< \dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.\Rightarrow0< x< \dfrac{2}{5}$ Vậy........... Chúc bạn học tốt!!!Câu trả lời: a, $x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x>\dfrac{2}{5}\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x< \dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.\Rightarrow0< x< \dfrac{2}{5}$ Vậy........... Chúc bạn học tốt!!!

Mashiro Shiina · Answer

$x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\Rightarrow x>\dfrac{2}{5}\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\Rightarrow x< \dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.$ $\Rightarrow0< x< \dfrac{2}{5}$ $\dfrac{x^2-1}{x^2}< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1>0\Rightarrow x^2>1\Rightarrow x>1\x^2< 0\Rightarrow x\in\varnothing\end{matrix}\right.\\left\{{}\begin{matrix}x^2-1< 0\Rightarrow x^2< 1\Rightarrow x< 1\x^2>0\Rightarrow x>0\end{matrix}\right.\end{matrix}\right.$ $\Rightarrow0< x< 1$Câu trả lời: $x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\Rightarrow x>\dfrac{2}{5}\end{matrix}\right.\\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\Rightarrow x< \dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.$ $\Rightarrow0< x< \dfrac{2}{5}$ $\dfrac{x^2-1}{x^2}< 0$ $\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1>0\Rightarrow x^2>1\Rightarrow x>1\x^2< 0\Rightarrow x\in\varnothing\end{matrix}\right.\\left\{{}\begin{matrix}x^2-1< 0\Rightarrow x^2< 1\Rightarrow x< 1\x^2>0\Rightarrow x>0\end{matrix}\right.\end{matrix}\right.$ $\Rightarrow0< x< 1$

Nguyễn Huy Tú · Answer

1, $x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\end{matrix}\right.$ ( loại ) hoặc $\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\end{matrix}\right.$ $\Rightarrow0< x< \dfrac{2}{5}$ Vậy... b, $\dfrac{x^2-1}{x^2}< 0$ $\Rightarrow\left\{{}\begin{matrix}x^2-1< 0\x^2>0\end{matrix}\right.$ hoặc $\left\{{}\begin{matrix}x^2-1>0\x^2< \end{matrix}\right.$ ( loại ) $\Rightarrow0< x^2< 1\Rightarrow0< x< 1$ Vậy...Câu trả lời: 1, $x^2-\dfrac{2}{5}x< 0$ $\Rightarrow x\left(x-\dfrac{2}{5}\right)< 0$ $\Rightarrow\left\{{}\begin{matrix}x< 0\x-\dfrac{2}{5}>0\end{matrix}\right.$ ( loại ) hoặc $\left\{{}\begin{matrix}x>0\x-\dfrac{2}{5}< 0\end{matrix}\right.$ $\Rightarrow0< x< \dfrac{2}{5}$ Vậy... b, $\dfrac{x^2-1}{x^2}< 0$ $\Rightarrow\left\{{}\begin{matrix}x^2-1< 0\x^2>0\end{matrix}\right.$ hoặc $\left\{{}\begin{matrix}x^2-1>0\x^2< \end{matrix}\right.$ ( loại ) $\Rightarrow0< x^2< 1\Rightarrow0< x< 1$ Vậy...

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