![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
6: =x^2-7xy+5xy-35y^2
=x(x-7y)+5y(x-7y)
=(x-7y)(x+5y)
7: =x^2-2xy-8xy+16y^2
=x(x-2y)-8y(x-2y)
=(x-2y)(x-8y)
8: =3x^2-6xy-4xy+8y^2
=3x(x-2y)-4y(x-2y)
=(x-2y)(3x-4y)
9: =4x^2+4xy+y^2-16y^2
=(2x+y)^2-16y^2
=(2x+y-4y)(2x+y+4y)
=(2x-3y)*(2x+5y)
10: =2(x^2+5xy+4y^2)
=2(x+y)(x+4y)
11: =5x(x+2y+y^2)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$M=x^2+8y^2-4xy+6x-16y+2019$
$=(x^2+4y^2-4xy)+4y^2+6x-16y+2019$
$=(x-2y)^2+6(x-2y)+4y^2-4y+2019$
$=[(x-2y)^2+6(x-2y)^2+9]+(4y^2-4y+1)+2009$
$=(x-2y+3)^2+(2y-1)^2+2009\geq 2009$
Vậy $M_{\min}=2009$. Giá trị này đạt tại $x-2y+3=0$ và $2y-1=0$ hay $(x,y)=(-2,\frac{1}{2})$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(9x^2+8y^2-12xy+6x-16y+10=0\)
\(\Rightarrow9x^2+8y^2-12xy+6x-16y=-10\)
\(=9x^2+2\left(4y^2-6xy+3x-8y\right)=-10\)
\(=9x^2+2\left[3x-6xy+4y\left(y-2\right)\right]\)
\(=9x^2+2\left[3x\left(1-2y\right)+4y\left(y-2\right)\right]\)
\(\Rightarrow\left\{{}\begin{matrix}9x^2=0\\\left\{{}\begin{matrix}1-2y=0\\y-2=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}y=\dfrac{1}{2}\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}y=\dfrac{1}{2}\\y=2\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=5x^2+y^2-4xy+4x-8y+1\)
\(=\left(y^2-4xy-8y\right)+5x^2+4x+1\)
\(=\left[y^2-2y\left(2x+4\right)+\left(2x+4\right)^2\right]+5x^2+4x+1-\left(2x+4\right)^2\)\(=\left(y-2x-4\right)^2+5x^2+4x+1-4x^2-16x-16\)\(=\left(y-2x-4\right)^2+\left(x^2-12x+36\right)-51\)
\(=\left(y-2x-4\right)^2+\left(x-6\right)^2-51\)
Với mọi giá trị của x ta có:
\(\left(y-2x-4\right)^2\ge0;\left(x-6\right)^2\ge0\)
\(\Rightarrow\left(y-2x-4\right)^2+\left(x-6\right)^2-51\ge-51\)
Vậy Min E = -51
Để E = 51 thì \(\left\{{}\begin{matrix}y-2x-4=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-12-4=0\\x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=16\\x=6\end{matrix}\right.\)
Lời giải:
\(E=100x^2+16y^2-8y+29=(10x)^2+(16y^2-8y+1)+28\)
\(=(10x)^2+(4y-1)^2+28\)
Vì \((10x)^2\geq 0; (4y-1)^2\geq 0, \forall x,y\in\mathbb{R}\)
\(\Rightarrow E\geq 0+0+28=28\)
Vậy \(E_{\min}=28\Leftrightarrow 10x=4y-1=0\Leftrightarrow x=0; y=\frac{1}{4}\)