Question

tim x va  y biet rang

a) x²+2y²+2xy-2y +1=0

b) x²+2y²+2xy -2x+2=0

Answer 1

......................?

mik ko biết

mong bn thông cảm 

nha ................

Answer 2

a) x²+2y²+2xy-2y+1=0

\(\Leftrightarrow\)(x²+2xy+y²)+(y²-2y+1)=0

\(\Leftrightarrow\)(x+y)²+(y-1)²=0

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

Vậy x=-1, y=1

Answer 3

a/ \(x^2+2y^2+2xy-2y+1=0\)

<=> \(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)

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

<=> \(\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-y\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

b/ \(x^2+2y^2+2xy-2x+2=0\)

<=> \(\left(x^2+2xy+y^2\right)+\left(2y-2x+2\right)=0\)

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

<=> \(\hept{\begin{cases}\left(x+y\right)^2=0\\2\left(y-x+1\right)=0\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\\y-x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\\y-x=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\left(1\right)\\x-y=1\left(2\right)\end{cases}}\)

Trừ (1) và (2)

=> \(2y=-1\)

<=> \(y=-\frac{1}{2}\)

<=> \(x=\frac{1}{2}\)(vì \(x+y=0\)<=> \(x=-y\))

Answer 4

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

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

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

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

Từ đó suy ra 2 trường hợp :

+ \(y-1=0\)                         + \(x+y=0\)              

\(\Rightarrow y=1\)                                \(\Rightarrow x=-y\)

                                                     \(\Rightarrow x=\left(-1\right)\)