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)
CM như kiểu là bé hoặc lớn hơn 0 vs mọi x,y á bạn thầy cô mk ghi đề vậy thì mk viết vậy thôi ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+y^2+2x-2xy+5-4y=0\)
\(\Leftrightarrow\left[y^2-2y\left(x+2\right)+\left(x+2\right)^2\right]+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(y-x-2\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-x-2=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
\(S=\left(x+2\right)^2+\left(y-1\right)^2=\left(1+2\right)^2+\left(3-1\right)^2\)
\(=3^2+2^2=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(x^2-8x+20\)
\(=x^2-8x+16+4\)
\(=\left(x-4\right)^2+4>0\forall x\)
b: Ta có: \(-x^2+6x-19\)
\(=-\left(x^2-6x+19\right)\)
\(=-\left(x^2-6x+9+10\right)\)
\(=-\left(x-3\right)^2-10< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x-3)(2x-7)=0
x=3 hoặc x=\(\dfrac{7}{2}\)
Vậy \(x\in\left\{3;\dfrac{7}{2}\right\}\)
Tìm GTNN của A=\(x^4-6x^3+12x^2-12x+2021\)
Giúp mk vs ạ mk đang cần gấp ai nhanh mk sẽ vote cho ạ :<
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Sửa:A=x^4-6x^3+13x^2-12x+2021\\ A=\left(x^4-6x^3+9x^2\right)+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x\right)^2+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x+2\right)^2+2017\ge2017\\ A_{min}=2017\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)