a) x^2 + 4y^2 + 6x - 12y + 18 = 0

<=>x²+6x+9+4y²-12y+9=0

<=>(x+3)²+(2y-3)²=0

<=>x+3=0 và 2y-3=0

<=>x=-3 và y=3/2

b) 5x^2 +9y^2 - 12xy - 6x +9 = 0

<=>x²-6x+9+4x²-12xy+9y²=0

<=>(x-3)²+(2x-3y)²=0

<=>x-3=0 và 2x-3y=0

<=>x=3 và 2.3-3y=0

<=>x=3 và y=2

Bài này giải rồi mà

${\mathrm{A=x}}^2-2x+2$

$\mathrm{=x2-2x+1+1}$

$\mathrm{=(x2-2x+1)+1}$

=(x-1)²+1

vì (x-1)²\(\ge0\forall x\)

=>(x-1)²+1\(\ge1\)

vậy A luôn dương với mọi x

B=x${}^{}+y^2+2x-4y+6$

=x²+2x+1+y²-4y+4+1

=(x²+2x+1)+(y²-4y+4)+1

=(x+1)²+(y-2)²+1

do (x+1)²\(\ge0\forall x\)

(y-2)²\(\ge0\forall y\)

=>(x+1)²+(y-2)²\(\ge0\)

=>(x+1)²+(y-2)²+1\(\ge1\)

=>B\(\ge1\)

vậy B luôn dương với mọi x;y

C= $x^2+y^2+z^2+4x-2y-4z+10$

$\mathrm{=x2+4x+4+y2-2y+1+z2-4z+4+1}$

=(x²+4x+4)+(y²-2y+1)+(z²-4z+4)+1

=(x+2)²+(y-1)²+(z-2)²+1

do (x+2)²\(\ge0\forall x\)

(y-1)²\(\ge0\forall y\)

(\(\)z-2)²\(\ge0\forall z\)

=>(x+2)²+(y-1)²+(z-2)²\(\ge0\)

=>(x+2)²+(y-1)²+(z-2)²+1\(\ge1\)

=>C\(\ge1\)

vậy C luôn dương với mọi x;y;z

bài 2: tìm x

a)\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2+y^2-2x+4y+1+4=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy x=1; y=-2

b)\(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(2x-3y\right)^2+\left(x-3\right)^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2.3-3.y=0\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)

Vậy x=2; y=3

x^2 -2x+2= x^2-2x+1+1=(x-1)^2+1> 0

vậy...

B= x^2+y^2+2x-4y+6

=(x^2+2x+1)+ (y^2-4y+2^2)+1

=(x+1)^2+(y-2)^2 >0

C, tương tự

a)

\(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-3y\right)^2\ge0\\\left(x-3\right)^2\ge0\end{cases}}\)nên

\(\Rightarrow\hept{\begin{cases}\left(2x-3y\right)^2=0\\\left(x-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

Vậy x=3 và y=2

b)

\(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)\(\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(x-5\right)^2\ge0\\\left(y-4\right)^2\ge0\end{cases}}\)nên

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(x-5\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\x-5=0\\y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=0\\x=5\\y=4\end{cases}}}\)( VÔ nghiệm vì \(x+y\ne0\))

Vậy không có giá trị x, y nào thỏa mãn đề bài

a , \(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow25x^2+45y^2-60xy-30x+45=0\)

\(\Leftrightarrow\left(5x\right)^2-2.5.\left(6y+3\right)+\left(6y+3\right)^2+9y^2-36y+36=0\)

\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y^2-4y+4\right)=0\)

\(\Leftrightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(5x-6y-3\right)^2\ge0\\9\left(y-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(5x-6y-3\right)^2+9\left(y-2\right)^2\ge0\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}5x-6y-3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy ...

      \(5x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow\left(4x^2+9y^2-12xy\right)+\left(x^2-6x+9\right)=0\)

\(\Rightarrow\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3y\\x=3\end{cases}\Rightarrow}\hept{\begin{cases}y=2\\x=3\end{cases}}}\)

\(5x^2+9y^2-12xy-6x+9=0\)

<=>  \(\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

<=>  \(\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

<=>  \(\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}}\)

<=>  \(\hept{\begin{cases}y=2\\x=3\end{cases}}\)

Vậy...