a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

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

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

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

\(\Rightarrow4x^2+4xy+y^2-4x-2y+1+y^2-2y+1=0\)

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

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

\(\Rightarrow4x^2+\left(y-1\right)^2=0\)

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

yiouoiyy

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

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

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

Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)

a: \(y^2-9-x^2+6x\)

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

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

=(y-x+3)(y+x-3)

b: \(25-4x^2-4xy-y^2\)

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

\(=25-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)

c: \(x^2-xz+4y^2-2yz+4xy\)

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

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

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

d: \(3x^2+6xy-48z^2+3y^2\)

\(=3\left\lbrack x^2+2xy+y^2-16z^2\right\rbrack\)

\(=3\left\lbrack\left(x+y\right)^2-\left(4z\right)^2\right\rbrack\)

=3(x+y+4z)(x+y-4z)

e: \(x^2-z^2+4y^2-4t^2-4xy+4zt\)

\(=x^2-4xy+4y^2-z^2+4zt-4t^2\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)

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

f: \(x^3+2x^2y+xy^2-16x\)

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

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

=x(x+y+4)(x+y-4)