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\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)(*)
Vì \(\left(x-1\right)\ge0;\left(y-3\right)^2\ge0;\left(z+1\right)^2\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\y=3\\z=-1\end{cases}}}\)
pt ⇔ ( 9x2 - 18x + 9 ) + ( y2 - 6y + 9 ) + ( 2z2 + 4z + 2 ) = 0
⇔ 9( x2 - 2x + 1 ) + ( y - 3 )2 + 2( z2 + 2z + 1 ) = 0
⇔ 9( x - 1 )2 + ( y - 3 )2 + 2( z + 1 )2 = 0
Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\2\left(z+1\right)^2\ge0\forall z\end{cases}}\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy
Bài 1a/
\(\frac{1}{1+x+xy}=\frac{xyz}{xyz+x+xy}=\frac{yz}{1+y+yz}\)
\(\frac{1}{1+z+xz}=\frac{y}{y+yz+xyz}=\frac{y}{1+y+yz}\)
Vậy \(M=\frac{1}{1+y+yz}+\frac{y}{1+y+yz}+\frac{yz}{1+y+yz}=1\)
Chiều về làm tiếp
Bài 1b:Lời giải này chủ yếu nhờ dự đoán trước Min là 2011/2012 đạt được khi x=2012
Ta có \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\ge\frac{2011x^2}{2012x^2}=\frac{2011}{2012}\)
Bài 2: Dùng phân tích thành bình phương
\(10x^2+y^2+4z^2+6x-4y-4xz+5=\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)\)
\(=\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}}\)
Bài 3:
a/\(pt\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\Leftrightarrow x=-6,x=5\)
b/ta phân tích vế trái thành:\(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
a, \(9x^2+y^2+2z^2-18x-6y+4z+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Mà \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy...
9x^2+ y^2 + 2z^2 - 18x + 4z - 6y + 20 = 0
<=>9x2-18x+9+y2-6y+9+2z2+4z+2=0
<=>(3x-3)2+(y-3)2+2.(z2+2z+1)=0
<=>(3x-3)2+(y-3)2+2.(z+1)2=0
<=>3x-3=0 và y-3=0 và z+1=0
<=>x=1 và y=3 và z=-1
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Rightarrow\left[\left(3x\right)^2-2.3x.3+9\right]+\left(y^2-2.y.3+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\left(3x-3\right)^2\ge0\) với mọi x
\(\left(y-3\right)^2\ge0\) với mọi y
\(2\left(z+1\right)^2\ge0\) với mọi z
\(\Rightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\) với mọi x, y, z
Mà \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-3=0\\y-3=0\\\left(z+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3\left(x-1\right)=0\\y=3\\z+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y=3\\z=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
Vậy x = 1 ; y = 3 ; z = -1