a) x² + 4y + 4y² + 26 - 10x = ( x² - 10x + 25 ) + ( 4y² + 4y + 1 ) = ( x - 5 )² + ( 2y + 1 )²

b) 4y² + 34 - 10x + 12y + x² = ( x² - 10x + 25 ) + ( 4y² + 12y + 9 ) = ( x - 5 )² + ( 2y + 3 )²

c) -10x + y² - 8y + x² + 41 = ( x² - 10x + 25 ) + ( y² - 8y + 16 ) = ( x - 5 )² + ( y - 4 )²

d) x² + 9y² - 12y + 29 - 10x = ( x² - 10x + 25 ) + ( 9y² - 12y + 4 ) = ( x - 5 )² + ( 3y - 2 )²

a) \(x^2+4y+4y^2+26-10x\)

\(=\left(x^2-10x+25\right)+\left(4y^2+4y+1\right)\)

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

b) \(4y^2+34-10x+12y+x^2\) đề ntn à?

\(=\left(4y^2+12y+9\right)+\left(x^2-10x+25\right)\)

\(=\left(2y-3\right)^2+\left(x-5\right)^2\)

c) \(-10x+y^2-8y+x^2+41\)

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

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

d) \(x^2+9y^2-12y+29-10x\)

\(=\left(x^2-10x+25\right)+\left(9y^2-12y+4\right)\)

\(=\left(x-5\right)^2+\left(3y-2\right)^2\)

\(x^2+9y^2-12y+29-10x\)

\(=\left(x^2-10x+25\right)+\left(9y^2-12y+4\right)\)

\(=\left(x^2-2.x.5+5^2\right)+\left[\left(3y\right)^2-2.3y.2+2^2\right]\)

\(=\left(x-5\right)^2+\left(3y-2\right)^2\)

1. Theo mình là sai đề, không biết có phải vậy không

2. (x^2 - 2.x.5 + 25) + (9y^2 - 2.3.2 +4) =0

(x-5)^2 + (3y-2)^2 = 0

TH1: (x-5)^2 = 0

  x-5=0

x=5

TH2:  (3y-2)^2 =0

3y -2=0

y=2/3

1. x²+y²-2x+4y+3=0

<=>(x²-2x+1)+(y²+4y+2)=0

<=>(x-1)²+(y+2)²=0

Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0}\)

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

\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

1) \(9x^2+y^2-10y-12x+29=0\)

\(\Leftrightarrow\left(9x^2-12x+4\right)+\left(y^2-10y+25\right)=0\)

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

ta có : \(\left(3x-2\right)^2\ge0\forall x\) và \(\left(y-5\right)^2\ge0\forall y\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=5\end{matrix}\right.\)

vậy \(x=\dfrac{2}{3};y=5\)

2) câu này đề sai rồi nha

3) \(x^2+29+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)+9=0\)

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

ta có : \(\left(x+4\right)^2\ge0\forall x\) và \(\left(3y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9\ge9>0\forall x;y\)

vậy phương trình vô nghiệm

Ta có: (x^2 + 9y^2 + 4- 6xy -12y+ 4x)+(x^2 -10x+25) =0

          (x-3y+2)^2 +(x-5)^2 =0

Vì vế trái luôn luôn lớn hơn hoặc bằng 0 với mọi x,y nên dấu"=" xảy ra khi:

   x-3y+2 =0 và x-5=0

   5-3y+2 =0 và x=5

   y=7/3 và x=5

Vậy x=5 và y=7/3.

Chúc bạn học tốt.

Cảm ơn bạn nhiều nha!

Bài 1:Tìm x,y biết:

a)\(x^2-6x+y^2+10y+34\)

=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)

=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)

=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)

Còn ý b,c,d,e làm tương tự ý a.