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a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
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 \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(\left(\frac{2x}{2x+y}-\frac{4x^2}{4x^2+2xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x\left(4x^2+2xy+y^2\right)}{\left(2x+y\right)\left(4x^2+2xy+y^2\right)}-\frac{4x^2\left(2x+y\right)}{\left(2x+y\right)\left(4x^2+2xy+y^2\right)}\right):\left(\frac{2}{\left(2x-y\right)\left(2x+y\right)}-\frac{2x+y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\frac{8x^3+4x^2y+2xy^2-8x^3-4x^2y}{\left(2x+y\right)\left(4x^2+2xy+y^2\right)}:\frac{2x-2x-y}{\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{2xy^2}{\left(2x+y\right)\left(4x^2+2xy+y^2\right)}.\frac{\left(2x-y\right)\left(2x+y\right)}{-y}\)
\(=\frac{4x^2-2xy^3}{-y\left(4x^2+2xy+y^2\right)}\)
Nếu chuyển \(4x^2+2xy+y^2\)thành \(4x^2-2xy+y^2\)thì nó sẽ dễ tính hơn nhiều ==
