![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
11c.
Từ đề bài ta có:
\(\left\{{}\begin{matrix}\dfrac{16a-b^2}{4a}=\dfrac{9}{2}\\16a+4b+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2b^2=-4a\\b=-4a-1\end{matrix}\right.\)
\(\Rightarrow2b^2-b=1\Leftrightarrow2b^2-b-1=0\Rightarrow\left[{}\begin{matrix}b=1\Rightarrow a=-\dfrac{1}{2}\\b=-\dfrac{1}{2}\Rightarrow a=-\dfrac{1}{8}\end{matrix}\right.\)
Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=-\dfrac{1}{2}x^2+x+4\\y=-\dfrac{1}{8}x^2-\dfrac{1}{2}x+4\end{matrix}\right.\)
4f.
Từ đề bài ta có:
\(\left\{{}\begin{matrix}1+b+c=0\\\dfrac{4c-b^2}{4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c=-b-1\\c=\dfrac{b^2}{4}-1\end{matrix}\right.\)
\(\Rightarrow\dfrac{b^2}{4}+b=0\)
\(\Rightarrow\left[{}\begin{matrix}b=0\Rightarrow c=-1\\b=-4\Rightarrow c=3\end{matrix}\right.\)
Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=x^2-1\\y=x^2-4x+3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
Gọi \(H\left(x;y\right)\) là toạ độ chân đường cao ứng với BC \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(x-1;y+2\right)\\\overrightarrow{BC}=\left(2;1\right)\end{matrix}\right.\)
Do AH vuông góc BC \(\Rightarrow\overrightarrow{AH}.\overrightarrow{BC}=0\)
\(\Rightarrow2\left(x-1\right)+y+2=0\Leftrightarrow y=-2x\)
\(\Rightarrow H\left(x;-2x\right)\Rightarrow\overrightarrow{BH}=\left(x+2;-2x-3\right)\)
Do H thuộc BC nên B, C, H thẳng hàng hay các vecto \(\overrightarrow{BC};\overrightarrow{BH}\) cùng phương
\(\Rightarrow\dfrac{x+2}{2}=\dfrac{-2x-3}{1}\Rightarrow x=\dfrac{8}{5}\Rightarrow y=-\dfrac{16}{5}\) \(\Rightarrow H\left(-\dfrac{8}{5};\dfrac{16}{5}\right)\)
\(\Rightarrow\overrightarrow{AH}=\left(-\dfrac{13}{5};\dfrac{26}{5}\right)\Rightarrow\left\{{}\begin{matrix}AH=\sqrt{\left(-\dfrac{13}{5}\right)^2+\left(-\dfrac{6}{5}\right)^2}=\dfrac{13\sqrt{5}}{5}\\BC=\sqrt{2^2+1^2}=\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{13}{2}\)
3.
Kẻ AD vuông góc BC tại D
\(\Rightarrow AD=BH=10\) ; \(BD=AH=4\)
\(tan\widehat{BAD}=\dfrac{BD}{AD}=\dfrac{2}{5}\Rightarrow\widehat{BAD}\approx21^048'5''\)
\(\Rightarrow\widehat{CAD}=60^0-\widehat{BAD}=38^011'55''\)
\(\Rightarrow CD=AD.tan\widehat{CAD}=7,87\left(m\right)\)
\(\Rightarrow BC=BD+CD=11,87\left(m\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
D E thuộc Ox \(\Rightarrow\) tọa độ E có dạng \(E\left(x;0\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{OE}=\left(x;0\right)\\\overrightarrow{OM}=\left(4;1\right)\end{matrix}\right.\)
Tam giác OEM cân tại O \(\Rightarrow OE=OM\)
\(\Rightarrow\sqrt{x^2+0^2}=\sqrt{4^2+1^2}\Rightarrow x^2=17\)
\(\Rightarrow x=\pm\sqrt{17}\Rightarrow\left[{}\begin{matrix}E\left(\sqrt{17};0\right)\\E\left(-\sqrt{17};0\right)\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}\overrightarrow{MA}=\left(a-4;-1\right)\\\overrightarrow{MB}=\left(-4;b-1\right)\end{matrix}\right.\)
Tam giác ABM vuông tại M \(\Rightarrow\overrightarrow{MA}.\overrightarrow{MB}=0\)
\(\Rightarrow-4\left(a-4\right)-1\left(b-1\right)=0\)
\(\Leftrightarrow4a+b-17=0\Rightarrow b=17-4a\)
Lại có \(S_{ABM}=\dfrac{1}{2}MA.MB=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{\left(b-1\right)^2+16}\)
\(=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{\left(16-4a\right)^2+16}=\dfrac{1}{2}\sqrt{\left(a-4\right)^2+1}.\sqrt{16\left[\left(a-4\right)^2+1\right]}\)
\(=2\left[\left(a-4\right)^2+1\right]\ge2\)
Dấu "=" xảy ra khi \(a-4=0\Rightarrow a=4\Rightarrow b=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: E thuộc Ox nên E(x;0)
O(0;0); M(4;1); E(x;0)
\(OM=\sqrt{\left(4-0\right)^2+\left(1-0\right)^2}=\sqrt{17}\)
\(OE=\sqrt{\left(x-0\right)^2+\left(0-0\right)^2}=\sqrt{x^2}=\left|x\right|\)
Để ΔOEM cân tại O thì OE=OM
=>\(\left|x\right|=\sqrt{17}\)
=>\(x=\pm\sqrt{17}\)