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Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}z=8\\x=-5\\y=-3\end{matrix}\right.\)
Vậy x = -5; y = -3; z = 8
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Rightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^3\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Mà \(\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=0\\x+5=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}z=-\left(x+y\right)\\x=-5\\y=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}z=8\\x=-5\\y=-3\end{matrix}\right.\)
a, B=x2+4xy+y2+x2-8x+16+2012
B=(x+y) 2+(x-4)2+2012
Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)
b làm tương tự
c, 9x2+6x+1+y2-4y+4+x2-4xz+4z2=0
(3x+1)2+(y-4)2+(x-2z)2=0
Vậy 3x+1=0 => x = -1/3
y-4=0 => y=4
x-2z=0 thế x=-1/3 ta được. -1/3-2z=0 => z = -1/6
Bạn nhớ ghi lại đề minh không ghi đề
a) \(B=2x^2+y^2+2xy-8x+2028\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)
\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)
b)\(C=x^2+5y^2+4xy+2x+2y-7\)
\(=\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+\left(y^2-2y+1\right)-9\)
\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)
\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)
a)\(2x^2+y^2+4x-2y-2xy+10=2x^2+y^2+4x-2y\left(x+1\right)+10\)
\(=y^2-2y\left(x+1\right)+2\left(x^2+2x+1\right)+8\)
\(=y^2-2y\left(x+1\right)+2\left(x+1\right)^2+8\)
\(=\left(y+x+1\right)^2+\left(x+1\right)^2+8\ge8\)
Dấu "=" xảy ra khi x=-1 và y=0
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
a/
\(\Leftrightarrow\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)-19=0\)
\(\Leftrightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2=19\)
Do 19 không thể phân tích thành tổng của 2 số chính phương nên pt vô nghiệm
b/
\(\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do x; y nguyên dương nên \(\left(2x+2y\right)^2>0\Rightarrow VT>0\)
Pt vô nghiệm
c/
\(\Leftrightarrow\left(x^2+4y^2+25-4xy+10x-20y+25\right)+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)
Do x;y;z nguyên dương nên \(\left|x+y+z\right|>0\Rightarrow VT>0\)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Do x;y;z nguyên dương nên vế phái luôn dương
Pt vô nghiệm