\(lim_{x\rightarrow+\infty}\left(\sqrt{ax^2+bx}-cx\right)=-2\)
\(\Leftrightarrow\sqrt{ax^2+bx}-cx=-2\left(x\rightarrow+\infty\right)\)(1)
\(\Leftrightarrow\frac{ax^2+bx-c^2x^2}{\sqrt{ax^2+bx}+cx}=-2\left(x\rightarrow+\infty\right)\)
\(\Leftrightarrow\frac{x\left(ax+b-c^2x\right)}{x\sqrt{a+\frac{b}{x}}+c}=-2\left(x\rightarrow+\infty\right)\)
\(\Leftrightarrow\frac{x\left(a-c^2\right)+b}{\sqrt{a}+c}=-2
\left(x\rightarrow+\infty\right)\)
\(\Rightarrow x\left(a-c^2\right)+b=-2\left(\sqrt{a}+c\right)
\left(x\rightarrow+\infty\right)\)
\(\Leftrightarrow a-c^2=\frac{-2\left(\sqrt{a}+c\right)-b}{x}\left(x\rightarrow+\infty\right)\)
\(\Rightarrow a-c^2=0\Leftrightarrow a=c^2\)
Mà \(c^2+a=18\)suy ra \(\hept{\begin{cases}c=\pm3\\a=9\end{cases}}\)
TH1: c=-3;a=9 thì (1) có giới hạn là vô cùng (loại)
TH2: c=3; a=9 thì (1) tương đương
\(\sqrt{9x^2+bx}-3x=-2\left(x\rightarrow+\infty\right)\)
\(\Leftrightarrow\frac{bx}{x\left(\sqrt{9+\frac{b}{x}}+3\right)}=-2\left(x\rightarrow+\infty\right)\)
\(\Leftrightarrow\frac{b}{6}=-2\Rightarrow b=-12\)
\(\Rightarrow a+b+5c=9-12+5.3=12\)
Giả sử cạnh hình vuông là a
\(AM=\frac{a}{2}\)
\(AN=\frac{3a\sqrt{2}}{4}\)
\(MN=\sqrt{\left(2-1\right)^2+\left(-1-2\right)^2}=\sqrt{10}\)
\(Cos\widehat{MAN}=\frac{AM^2+AN^2-MN^2}{2AM.AN}\)
\(\Leftrightarrow\frac{\sqrt{2}}{2}=\frac{\frac{1}{4}a^2+\frac{9}{8}a^2-10}{2.\frac{1}{2}a.\frac{3\sqrt{2}}{4}a}\Rightarrow a=4\)
Giả sử CD: \(\left(d\right):y=ax+b\)
MN cắt CD tại K \(\Rightarrow K\in\left(d\right)\)
Ta có:
\(\Delta MNA\infty\Delta KNC\)
\(\Rightarrow\frac{MN}{NK}=\frac{AN}{NC}=3\)
\(\Leftrightarrow\overrightarrow{MN}=3\overrightarrow{NK}\Rightarrow K\left(\frac{7}{3};-2\right)\)
Do \(K\in\left(d\right)\Rightarrow7a+3b=-6\)(1)
Viết lại \(\left(d\right):ax-y+b=0\)
\(d_{\left(M,\left(d\right)\right)}=4\Rightarrow\frac{\left|a-2+b\right|}{\sqrt{a^2+1}}=4\Leftrightarrow\left(a-2+b\right)^2=16\left(a^2+1\right)\)(2)
Từ (1) và (2) \(\hept{\begin{cases}a=0\\b=-2\end{cases}}\)hoặc \(\hept{\begin{cases}a=\frac{3}{4}\\b=-\frac{15}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(d\right):y+2=0\\\left(d\right):3x-4y-15=0\end{cases}}\)
