\(\left(x+2\right)\left(5-2x\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2\ge0\\5-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2\le0\\5-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-2\\x\le\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-2\\x\ge\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\) (Loại TH2) \(\Leftrightarrow\dfrac{5}{2}\ge x\ge-2\)
\(d.\sqrt{\dfrac{2x+3}{7-x}-1}\)
\(\dfrac{2x+3}{7-x}-1\ge0\)
\(\Leftrightarrow\dfrac{2x+3-7+x}{7-x}\ge0\)
\(\Leftrightarrow\dfrac{3x-4}{7-x}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-4\ge0\\7-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-4\le0\\7-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{4}{3}\\x< 7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{4}{3}\\x>7\end{matrix}\right.\end{matrix}\right.\) (loại TH2)
\(\Leftrightarrow7>x\ge\dfrac{4}{3}\)
Bài 1:
a.Áp dụng BĐT Cô- si, ta có:
\(\sqrt{3}+\sqrt{5}\ge2\sqrt{\sqrt{3}.\sqrt{5}}=2\sqrt{\sqrt{15}}\) (1)
\(\sqrt{6}+\sqrt{2}\ge2\sqrt{\sqrt{6}.\sqrt{2}}=2\sqrt{\sqrt{12}}\) (2)
Từ (1) và (2) suy ra:
\(\sqrt{3}+\sqrt{5}>\sqrt{6}+\sqrt{2}\)
b. Ta có: \(\sqrt{4}+\sqrt{3}=2+\sqrt{3}>2+1=3\)
Suy ra: \(\sqrt{4}+\sqrt{3}>3\)
Bài 2:
a. \(\sqrt{\dfrac{2x+1}{4-3x}}\ge0\)
* TH1: \(2x+1\ge0\) và \(4-3x\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\) và \(x\le\dfrac{4}{3}\)
\(\Rightarrow\dfrac{4}{3}\ge x\ge-\dfrac{1}{2}\)
* TH2: \(2x+1\le0\) và \(4-3x\le0\)
\(\Rightarrow x\le-\dfrac{1}{2}\) và \(x\ge\dfrac{4}{3}\) (vô lý)
Vậy: \(\dfrac{4}{3}\ge x\ge-\dfrac{1}{2}\)
\(b.\sqrt{\dfrac{3}{7}-\dfrac{2}{9}x}\)
\(\dfrac{3}{7}-\dfrac{2}{9}x\ge0\Rightarrow x\le\dfrac{27}{14}\)
\(c.\sqrt{\left(x+2\right)\left(5-2x\right)}\)
\(\left(x+2\right)\left(5-2x\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2\ge0\\5-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2\le0\\5-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\ge-2\\x\le\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-2\\x\ge\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{5}{2}\ge x\ge-2\)
\(c.\sqrt{\left(x+2\right)\left(5-2x\right)}\)
\(\left(x+2\right)\left(5-2x\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2\ge0\\5-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2\le0\\5-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\ge-2\\x\le\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-2\\x\ge\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{5}{2}\ge x\ge-2\)
