1.
a, \(\left(2x-5\right)\left(x^2-x-6\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5< 0\\x^2-x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5>0\\x^2-x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\\left[{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\-2< x< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -2\\\dfrac{5}{2}< x< 3\end{matrix}\right.\)
Vậy ...
b, \(\dfrac{3x+4}{x^2-3x+5}< 0\)
Vì \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\) nên bất phương trình tương đương:
\(3x+4>0\Leftrightarrow x>-\dfrac{4}{3}\)
c, \(\left|3x+7\right|>2x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+7< -2x-3\\3x+7>2x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -2\\x>-4\end{matrix}\right.\)
\(\Rightarrow x\in R\)
2.
Phương trình có hai nghiệm phân biệt khi:
\(\Delta'=\left(m-1\right)^2-\left(2m+1\right)=m^2-4m>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m< 0\\m>4\end{matrix}\right.\)
5.
a, \(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-2;-1\right)\\\overrightarrow{BC}=\left(4;5\right)\\\overrightarrow{CA}=\left(-2;-4\right)\end{matrix}\right.\)
Phương trình đường thẳng AB:
\(\dfrac{x-3}{-2}=\dfrac{y-2}{-1}\Leftrightarrow x-2y+1=0\)
Phương trình đường thẳng BC:
\(\dfrac{x-1}{4}=\dfrac{y-1}{5}\Leftrightarrow5x-4y-1=0\)
Phương trình đường thẳng CA:
\(\dfrac{x-5}{-2}=\dfrac{y-6}{-4}\Leftrightarrow2x-y-4=0\)
\(\Leftrightarrow x+2y-17=0\)
b, \(\left\{{}\begin{matrix}AH\perp BC\\A=\left(3;2\right)\end{matrix}\right.\Rightarrow4x+5y-22=0\left(AH\right)\)
M có tọa độ: \(\left\{{}\begin{matrix}x_M=\dfrac{x_B+x_C}{2}=3\\y_M=\dfrac{y_B+y_C}{2}=\dfrac{7}{2}\end{matrix}\right.\Rightarrow M=\left(3;\dfrac{7}{2}\right)\)
Ta có \(\overrightarrow{AM}=\left(0;\dfrac{3}{2}\right)\)
\(\Rightarrow\) Phương trình tham số của AM: \(\left\{{}\begin{matrix}x=3\\y=2+\dfrac{3}{2}t\end{matrix}\right.\)
