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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
1)
a) \(x^2-4x+y^2+2y+5=0\)
\(\left(x^2-4x+4\right)+\left(y^2+2x+1\right)=0\)
\(\left(x-2\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
b) \(x^2+2y^2+2xy-2y+1=0\)
\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c) \(x^2+2y^2+2xy=2y-2\)
\(x^2+2y^2+2xy-2y+2=0\)
\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+1=0\)
\(\left(x+y\right)^2+\left(y-1\right)^2+1=0\)
Ta thấy \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\\1>0\end{matrix}\right.\)
\(\Rightarrow\) \(\left(x+y\right)^2+\left(y-1\right)^2+1>0\)
Vậy ko có x và y thoả mãn bài toán
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=0\\y=1\end{cases}\Rightarrow}x=-1}\)
Vậy x=-1 ; y=1
\(x^2+y^2+26+10x+2y=0\)
\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)( do \(\left(x+5\right)^2\ge0;\left(y+1\right)^2\ge0\))
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
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
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
Ta có : x2 - 4x + y2 + 2y + 5 = 0
<=> (x2 - 4x + 4) + (y2 + 2y + 1) = 0
<=> (x - 2)2 + (y + 1)2 = 0
Mà (x - 2)2 \(\ge0\forall x\)
(y + 1)2 \(\ge0\forall x\)
Nên \(\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\y=-0\end{cases}}\)
còn 2 bài nữa giúp mik đi