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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 \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
a) \(2x^2+y^2+2xy+10x+25=0\)
\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)
\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)
Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)
b)\(x^2+3y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2y^2-2y+\frac{1}{2}\right)+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)
Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)
nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)
Mà\(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)
nên pt vô nghiệm