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\(5x^2+2y^2-6xy+16x-8y+16=0\)
\(\Rightarrow10x^2+4y^2-12xy+32x-16y+32=0\)
\(\Rightarrow\left(9x^2-12xy+4y^2\right)+\left(24x-16y\right)+16+\left(x^2+8x+16\right)=0\)
\(\Rightarrow\left(3x-2y\right)^2+2.\left(3x-2y\right).4+4^2+\left(x+4\right)^2=0\)
\(\Rightarrow\left(3x-2y+4\right)^2+\left(x+4\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x-2y+4=0\\x+4=0\end{cases}\Rightarrow}\hept{\begin{cases}-12-2y+4=0\\x=-4\end{cases}\Rightarrow\hept{\begin{cases}y=-4\\x=-4\end{cases}}}\)
Vậy \(x=y=-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình làm câu đầu tượng trưng thui nhé, 2 câu sau tương tự vậy !!!!!!
a) pt <=> \(x^2-2xy+2y^2-2x-2y+5=0\)
<=> \(\left(x-y-1\right)^2+y^2-4y+4=0\)
<=> \(\left(x-y-1\right)^2+\left(y-2\right)^2=0\) (1)
TA LUÔN CÓ: \(\left(x-y-1\right)^2;\left(y-2\right)^2\ge0\forall x;y\)
=> \(\left(x-y-1\right)^2+\left(y-2\right)^2\ge0\) (2)
TỪ (1) VÀ (2) => DẤU "=" SẼ PHẢI XẢY RA <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)
VẬY \(\left(x;y\right)=\left(3;2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)xy+3x-2y=11\)
\(\Leftrightarrow xy+3x-2y-6=5\)
\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)
\(b)2x^2-2xy+x-y=12\)
\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)
\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)
\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)
\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
Vì 2x+1 luôn lẻ
\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
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\(\left(\dfrac{x}{x+2y}-\dfrac{x+2y}{2y}\right)\left(\dfrac{x}{x-2y}-1+\dfrac{8y^3}{8y^3-x^3}\right)=\dfrac{2xy-\left(x+2y\right)^2}{2y\left(x+2y\right)}\left(\dfrac{2y}{x-2y}+\dfrac{8y^3}{\left(2y-x\right)\left(4y^2+2yx+x^2\right)}\right)=\dfrac{-\left(x^2+2xy+4y^2\right)}{2y\left(x+2y\right)}\cdot\dfrac{2y\left(4y^2+2yx+x^2\right)-8y^3}{\left(x-2y\right)\left(x^2+2xy+4y^2\right)}=\dfrac{-\left(x^2+2xy+4y^2\right)2y\left(4y^2+2xy+x^2-4y^2\right)}{2y\left(x+2y\right)\left(x-2y\right)\left(x^2+2x+4y^2\right)}=\dfrac{-\left(x^2+2xy\right)}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{x}{2y-x}\)
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Ta có : \(5x^2+2y^2-6xy+16x-8y+16=0\)
\(\Leftrightarrow10x^2+4y^2-12xy+32x-16y+32=0\)
\(\Leftrightarrow\left(4y^2-2.2y.3x+9x^2\right)+x^2+32x-16y+32=0\)
\(\Leftrightarrow\left(2y-3x\right)^2-2.4.\left(2y-3x\right)+16+x^2+8x+16=0\)
\(\Leftrightarrow\left(2y-3x-4\right)^2+\left(x+4\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2y-3x-4\right)^2=0\\\left(x+4\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-4\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left(-4,4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x-2y-2)2+(y-6)2 =39-2A
A=< 39/2, max A là 39/2 khi x =14 và y =6
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a: x^2-2x+y^2-8y+17=0
=>x^2-2x+1+y^2-8y+16=0
=>(x-1)^2+(y-4)^2=0
=>x=1 và y=4
b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0
=>(2x-y)^2+(y+2)^2=0
=>y=-2 và x=-1
x = 4 nha.
x = 4 nha.