x^2+10x +26+y^2 +2y
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)
<=> [ (x^2+2xy+y^2)+ 2.(x+y).5 +25 ] + (y^2+2y+1)=0
<=> (x+y+5)^2 + (y+1)^2 = 0
<=> x+y+5 = 0 và y+1 = 0
<=> x=-4 và y=-1
Ta có: x2+2y2+2xy+10x+12y+26=0
=> (x2+2xy+y2)+(10x+10y)+25+(y2+2y+1)=0
=> (x+y)2+10(x+y)+25+(y2+2y+1)=0
=> (x+y+5)2+(y+1)2=0
=> (x+y+5)2=(y+1)2=0
=> x+y+5=y+1=0
(+) y+1=0=> y=-1
(+) x+y+5=0 mà y=-1=> x-1+5=0
=> x+4=0=> x=-4
Vậy (x,y)=(-4;-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =x^2-10x+25+y^2+2y+1
=(x-5)^2+(y+1)^2>=0
Dấu = xảy ra khi x=5 và y=-1
b: x^2-3x-2
=x^2-3x+9/4-17/4
=(x-3/2)^2-17/4>=-17/4
Dấu = xảy ra khi x=3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+10x+26+y^2+2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy \(x=-5\)và \(y=-1\)
\(x^2+10x+26+y^2+2y=0\)
\(\Leftrightarrow x^2+10x+25+y^2+2y+1=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy..............
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+26+10x-2y=0\)
\(\left(x^2+10x\right)+\left(y^2-2y\right)+26=0\)
\(\left(x^2+2.x.5+5^2\right)+\left(y^2-2.y.1+1^2\right)=0\)
\(\left(x+5\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\y-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\y=1\end{array}\right.\)
j) \(x^2+y^2+26+10x-2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\begin{cases}x+5=0\\y-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-5\\y=1\end{cases}\)
Vậy x=-5; y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+10x+26+y^2+2y\)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
\(\left(x+y+4\right)\left(x+y-4\right)\)
\(=\left(x+y\right)^2-16\)
\(=x^2+y^2+2xy-16\)
a, =(x^2 +10x+25) +(y^2 +2y+1)
= (x+5)^2 +(y+1)^2
b, =(x+y)^2 -4^2
= x^2 + 2xy+ y^2 -16
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 + 10x + 26 + y2 + 2y
= x2 + 10 + 25 + 1 + y2 + 2y
= (x2 + 10x + 25) + (y2 + 2y + 1)
= (x + 5)2 + (y + 1)2
x2 - 2xy + 2y2 + 2y + 1
= x2 - 2xy + y2 + y2 + 2y + 1
= (x2 - 2xy + y2) + (y2 + 2y + 1)
= (x - y)2 + (y + 1)2
4x2 + 2z2 - 4xz - 2z + 1
= 4x2 + z2 + z2 - 4xz - 2z + 1
= (4x2 - 4xz + z2) + (z2 - 2z + 1)
= (2x + z)2 + (z - 1)2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=4x^2+4x+8\)
\(=4\left(x^2+x+\dfrac{1}{4}\right)+7\)
\(=4\left(x+\dfrac{1}{2}\right)^2+7\ge7\forall x\)
Vậy Min A = 7 khi \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)
\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(=4x^2-4x+1+x^2+4x+4\)
\(=5x^2+5\)
Vậy Min B = 5 khi \(x=0\)
\(C=x^2+10x+26+y^2+2y+2020\)
\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)+2020\)
\(=\left(x+5\right)^2+\left(y+1\right)^2+2020\ge2020\forall x\)
Vậy Min C = 2020 khi \(\left\{{}\begin{matrix}x+5=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-1\end{matrix}\right.\)
\(=x^2+10x+25+y^2+2y+1\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)