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\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
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 \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
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