\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

Ta có:

\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)

Không tồn tại x,y,z thỏa mãn đề bài

\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=1\\x=4\\y=3\end{cases}}\)

Vậy \(x=4\), \(y=3\), \(z=1\)

Ta có x² - 2xy + 2y² -2x + 6y+5 =0

<=> (x² - 2xy + y²) - (2x - 2y) + (y² + 4y + 4) + 1 = 0

<=> [(x - y)² - 2(x - y) + 1] + (y + 2)² = 0

<=> (x - y - 1)² + (y + 2)² = 0

<=> \(\hept{\begin{cases}x-y-1=0\\2\:+y=0\end{cases}}\)

<=> (x; y) = (-1; -2)

x² - 2xy + 2y² + 2x - 6y + 4 = 0

<=> [ ( x² - 2xy + y² ) + 2( x - y ) + 1 ] + ( y² - 4y + 4 ) - 1 = 0

<=> [ ( x - y )² + 2( x - y ) + 1 ] + ( y - 2 )² - 1 = 0

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

<=> ( x - y + 1 )² + ( y - 2 )² = 1

Nhận thấy rằng VT là tổng của hai bình phương 

=> VP cũng phải là tổng của hai bình phương

Ta có : 1 = 1² + 0²

               = (-1)² + 0²

Ta xét 4 trường hợp sau :

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

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

3. \(\hept{\begin{cases}\left(x-y+1\right)^2=0^2\\\left(y-2\right)^2=1^2\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)

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

Vậy ( x ; y ) = { ( 0 ; 2 ) , ( 2 ; 2 ) , ( 2 ; 3 ) , ( 0 ; 1 ) }

\(x^2-2xy+y^2+2x-6y+4=0\)

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

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

Mà \(x;y\in Z\); \(\left(x-y+1\right)^2\ge0;\left(y-2\right)^2\ge0\)

pt <=> \(\orbr{\begin{cases}\left(x-y+1\right)^2=0\\\left(y-2\right)^2=1\end{cases}}\) hoặc \(\orbr{\begin{cases}\left(x-y+1\right)^2=1\\\left(y-2\right)^2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x-y=-1\\y=3\end{cases}}\) hoặc \(\orbr{\begin{cases}x-y=0\\y=2\end{cases}}\)

<=> x = 2 ; y = 3 hoặc x = y = 2 ( tm x ; y thuộc Z )

Vậy các cặp số x ; y thỏa mãn pt trên là : ( 2 ; 3 ) ; ( 2 ; 2 ) 

Thiếu v:

pt <=> \(\orbr{\begin{cases}\left(x-y+1\right)^2=0\\\left(y-2\right)^2=1\end{cases}}\) hoặc \(\orbr{\begin{cases}\left(x-y+1\right)^2=1\\\left(y-2\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-y=-1\\y=3;y=1\end{cases}}\)hoặc \(\orbr{\begin{cases}x-y=0;x-y=-2\\y=2\end{cases}}\)

<=> x = 2 ; y = 3 hoặc x = 0 ; y = 1 hoặc x = y = 2 hoặc x = 0 ; y = 2

2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0  

(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0  

( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0

( x + y + z)2 = 0 ;

( x + 5)2 = 0 ;

(y + 3)2 = 0

vậy x = - 5 ; y = -3; z = 8 

Tìm x, y, z biết rằng: 2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0

                                Giải

2x2 + 2y2 + z2 + 2xy + 2xz + 2yz + 10x + 6y + 34 = 0 

(x2 + y2 + z2 + 2xy + 2xz + 2yz) + (x2 + 10x + 25) + (y2+ 6y + 9) = 0

 ( x + y + z)2 + ( x + 5)2 + (y + 3)2 = 0 

( x + y + z)2 = 0 ; ( x + 5)2 = 0 ; (y + 3)2 = 0

x = - 5 ; y = -3; z = 8 

Phần trả lời của mình cũng giống như Arcobaleno vậy

Ta có: \(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow\left(x^2-8x+16\right)+\left(y^2-6y+9\right)+\left(x^2-2xy+y^2\right)-2\left(x-y\right)z+z^2=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(y-3\right)^2+\left[\left(x-y\right)^2-2\left(x-y\right)z+z^2\right]=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y-3\right)^2=0\\\left(x-y-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=1\end{cases}}\)

\(2x^2+2y^2+z^2+2xy+2yz+2xz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

\(\Rightarrow x=-5,y=-3,z=8\)