a/ \(\left\{{}\begin{matrix}9x-15>4x+13\\19x-5x< 7+3x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x>28\\11x< 7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\frac{28}{5}\\x< \frac{7}{11}\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại x thỏa mãn
Vậy BPT vô nghiệm
b/ Ko nhìn thấy
c/ \(\left\{{}\begin{matrix}\frac{x}{2}+\frac{1}{2}-\frac{x}{3}-\frac{2}{3}< 2+\frac{x}{6}\\\frac{3x}{4}+\frac{5}{4}-1< \frac{x}{3}-\frac{2}{3}+x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}< 2\left(luôn-đúng\right)\\\frac{7x}{12}>\frac{11}{12}\end{matrix}\right.\) \(\Rightarrow x>\frac{11}{7}\)
d/ \(\left\{{}\begin{matrix}5x-2< 4x+5\\x^2< x^2+4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 7\\4x>-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< 7\\x>-1\end{matrix}\right.\)
\(\Rightarrow-1< x< 7\)
e/ \(\left\{{}\begin{matrix}x^2+2x+1>x^2-3x+5\\x^3-6x^2-7x+5< x^3-6x^2+12x-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x>4\\19x>13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>\frac{4}{5}\\x>\frac{13}{19}\end{matrix}\right.\) \(\Rightarrow x>\frac{4}{5}\)
f/ \(\left\{{}\begin{matrix}\frac{\left(x-1\right)^2}{x-2}\ge0\\2x>4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x>2\end{matrix}\right.\\x>2\end{matrix}\right.\) \(\Rightarrow x>2\)
g/ \(\left\{{}\begin{matrix}\frac{2}{2x-1}-\frac{1}{3-x}\le0\\\left|x\right|< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{7-4x}{\left(2x-1\right)\left(3-x\right)}\le0\\-1< x< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< \frac{1}{2}\\\frac{7}{4}\le x< 3\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\) \(\Rightarrow-1< x< \frac{1}{2}\)
h/ \(\left\{{}\begin{matrix}\frac{2x-1}{x-2}\ge0\\\frac{\left(x-1\right)\left(x+1\right)}{x-2}\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le\frac{1}{2}\\x>2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\1\le x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x\le-1\)
i/ \(\left\{{}\begin{matrix}\frac{5x-\sqrt{2}}{x\sqrt{2}-3}>0\\x^2< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< \frac{\sqrt{2}}{5}\\x>\frac{3\sqrt{2}}{2}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\) \(\Rightarrow-1< x< \frac{\sqrt{2}}{5}\)
j/ \(\left\{{}\begin{matrix}\frac{2x+3}{x-1}-1\ge0\\\frac{2\left(x-2\right)\left(x+2\right)}{x-1}\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x+4}{x-1}\ge0\\\frac{2\left(x-2\right)\left(x+2\right)}{x-1}\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x\le-4\end{matrix}\right.\\\left[{}\begin{matrix}x\le-2\\1< x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\le-4\\1< x\le2\end{matrix}\right.\)
