<=> [ (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)

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

\(x^2+10x+26+y^2+2y=0\)

\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)

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..............

\(x^2+y^2+26+10x-2y=0\)

\(\left(x^2+10x\right)+\left(y^2-2y\right)+26=0\)

\(\left(x^2+2.x.5+5^2\right)+\left(y^2-2.y.1+1^2\right)=0\)

\(\left(x+5\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\y-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\y=1\end{array}\right.\)

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

\(x^2+10x+26+y^2+2y\)

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

\(\left(x+y+4\right)\left(x+y-4\right)\)

\(=\left(x+y\right)^2-16\)

\(=x^2+y^2+2xy-16\)

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

x²+10x+26+y²+2y=(x+5)²+(y+1)²

(x+y+4)(x+y-4)=(x+y)²-16

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

x^2 +10x +26 +y^2 +2y=(x^2+10x+25)+(y^2+2y+1)

=(x+5)^2+(y+1)^2

#)Giải :

\(x^2+10x+26+y^2+2y=\left(x^2+10+25\right)+\left(y^2+2y+1\right)=x^2+2.5.x+5^2+y^2+2y+1=\left(x+5\right)^2+\left(y+1\right)^2\)

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

\(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)

\(x^2+10x+26+y^2+2y=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=\left(x+5\right)^2+\left(y+1\right)^2\)

Đề? ^~~^

\(A=4x^2+4x+8\)

\(=4\left(x^2+x+\dfrac{1}{4}\right)+7\)

\(=4\left(x+\dfrac{1}{2}\right)^2+7\ge7\forall x\)

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 \(\left\{{}\begin{matrix}x+5=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-1\end{matrix}\right.\)