a) \(x^2+4y^2-6x-4y+10=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(2y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-3=0\\2y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

b) \(2x^2+y^2+2xy-10x+25=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x-5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-5\\x=5\end{cases}}\)

c) \(x^2+2xy+4x-4y-2xy+5=0\)

\(\Leftrightarrow x^2-4x-4y+5=0\)

Xem lại đề câu c).

a) x² + 4y² - 6x - 4y + 10 = 0

<=> x² - 6x + 9 + 4y² - 4y + 1 = 0

<=> ( x - 3 )² + ( 4y - 1 )² = 0

<=> \(\hept{\begin{cases}x-3=0\\4y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{4}\end{cases}}\)

b) 2x² + y² + 2xy - 10x + 25 = 0

<=> x² + 2xy + y² + x² - 10x + 25 = 0

<=> ( x + y )² + ( x - 5 )² = 0

<=> \(\hept{\begin{cases}x+y=0\\x-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-5\\x=5\end{cases}}\)

c) Xem lại đề 

điều kiện: x + y khác 0, z khác 0, chắc vậy :v

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

= \(\frac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x+y-z\right)\left(x+y\right)}\)

= \(\frac{x+y+z}{x+y}\)

rút gọn phân thức giùm mk nha

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

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

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

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

=x^4-y^4

k cho mk nhe mấy bạn

a) \(2x^2+y^2+4x-2y-2xy+10\)

\(=x^2+x^2+y^2+4x-2y-2xy+4+6\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)-2\left(y-3\right)\)

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

.......................chắc không phải cách làm này đâu!

b) \(5x^2+y^2+2xy-4x\)

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

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

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

a, \(2x^2\)+\(y^2\)+\(4x-2y-2xy+10\)\(=y^2\)\(-x^2\)\(-1+2x-2y-2xy+3x^2+2x+11\)\(=\left(y-x-1^{ }\right)^2\)\(+3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{32}{3}\)\(=\left(y-x-1\right)^2+3\left(x+\frac{1}{3}\right)^2+\frac{32}{3}\)\(\ge\frac{32}{3}\)

VẬY GTNN CỦA BIỂU THỨC \(=\frac{32}{3}\)KHI \(y-x-1=0;x+\frac{1}{3}=0\Rightarrow x=\frac{-1}{3};y=\frac{2}{3}\)

\(\dfrac{x^2-2xy+y^2+2z\left(x-y\right)+z^2}{\left(x-y\right)^2-z^2}=\dfrac{\left(x-y\right)^2+2z\left(x-y\right)+z^2}{\left(x-y-z\right)\left(x-y+z\right)}\)

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

\(\frac{x^2+y^2+z^2-2xy-2yz+2zx}{x^2-2xy+y^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\frac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}=\frac{x-y+z}{x-y-z}\)

a)\(M=x^2-2xy+2y^2-4y+2016\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)

Dấu = khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-2=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=y\\y=2\end{cases}\)\(\Leftrightarrow x=y=2\)

Vậy Min_M=2012 khi x=y=2

b)\(N=x^2-2xy+2x+2y^2-4y+2016\)

\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-2y+1\right)+2014\)

\(=\left(x-y+1\right)^2+\left(y-1\right)^2+2014\ge2014\)

Dấu = khi \(\begin{cases}\left(x-y+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y+1=0\\y-1=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x-y+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x-1+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\y=1\end{cases}\)

Vậy Min_N=2014 khi x=0;y=1