![](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)
C,D thuộc (P) nên \(C\left(x_1;x_1^2-4x_1-5\right);D\left(x_2;x_2^2-4x_2-5\right)\)
ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
\(\overrightarrow{AB}=\left(3;-4\right);\overrightarrow{DC}=\left(x_1-x_2;x_1^2-4x_1-5-x_2^2+4x_2+5\right)\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1^2-x_2^2-4x_1+4x_2=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\\left(x_1-x_2\right)\left(x_1+x_2-4\right)=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2-4=-\dfrac{4}{x_1-x_2}=-\dfrac{4}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2=-\dfrac{4}{3}+4=\dfrac{12}{3}-\dfrac{4}{3}=\dfrac{8}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x_1=3+\dfrac{8}{3}=\dfrac{17}{3}\\x_1+x_2=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{17}{6}\\x_2=\dfrac{8}{3}-\dfrac{17}{6}=-\dfrac{1}{6}\end{matrix}\right.\)
Khi x=17/6 thì \(y=x^2-4x-5=\left(\dfrac{17}{6}\right)^2-4\cdot\dfrac{17}{6}-5=-\dfrac{299}{36}\)
Khi x=-1/6 thì \(y=\left(-\dfrac{1}{6}\right)^2-4\cdot\dfrac{-1}{6}-5=\dfrac{1}{36}+\dfrac{2}{3}-5=-\dfrac{155}{36}\)
Vậy: \(C\left(\dfrac{17}{6};-\dfrac{299}{36}\right);D\left(-\dfrac{1}{6};-\dfrac{155}{36}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Tọa độ điểm D là:
\(\left\{{}\begin{matrix}x_D=\dfrac{1-1}{2}=0\\y_D=\dfrac{-2+\left(-2\right)}{2}=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\left\{x\in N^+|-3< x\le2\right\}\)
\(\Rightarrow A=\left\{-2;-1;0;1;2\right\}\)
\(\Rightarrow A=D=\left\{-2;-1;0;1;2\right\}\)
Vậy chọn C
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt \(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)
Ta có \(A=(a-\frac{ab^2}{1+b^2})+(b-\frac{bc^2}{1+c^2})+(c-\frac{ca^2}{1+a^2})=3-\left ( \frac{ab^2}{1+b^2}+\frac{bc^2}{1+c^2}+\frac{ca^2}{1+a^2} \right )\)
Áp dụng bất đẳng thức AM-GM:
\(A\geq 3-\left ( \frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{3a} \right )=3-\frac{1}{2}(ab+bc+ac)\)
Cũng theo AM-GM
\(9=(a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 3\)
\(\Rightarrow A\geq 3-\frac{3}{2}=\frac{3}{2}\)
Dấu $=$ xảy ra khi \(a=b=c=1\)
B.2 nha
cho mình nhoa
HT
B .2 NHA