\(2x^2-2xy+2y^2-6x-6y+18=0\)

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

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

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

Mà \(\left(x-y\right)^2\ge0;\left(y-3\right)^2\ge0;\left(x-3\right)^2\ge0\)

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

Bài 1:

a) x( x - y) + x - y = (x - y)(x + 1)

b) 2x + 2y - x( x + y) = ( 2x + 2y) - x( x + y)

= 2( x + y ) - x( x + y ) = ( x + y )(2 - x )

c) 5x² - 5xy - 10x + 10y = ( 5x² - 5xy ) - ( 10x - 10y)

= 5x( x - y ) - 10( x - y ) = ( x - y )(5x - 10 )

= 5( x - y )( x - 2 )

d) 4x² + 6xy - 3x - 6y = Mình ko làm được!!! bạn chép có sai đề không

Bài 2:

x ( 2x - 7) - 4x + 14 = 0

⇒ 2x² - 7x - 4x + 14 = 0 ⇒ ( 2x² - 4x ) - ( 7x - 14 ) = 0

⇒ 2x( x - 2 ) - 7(x - 2) = 0

⇒ (x - 2)(2x - 7) = 0

⇒ \(\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x = 2; x = \(\dfrac{7}{2}\)

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