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Lời giải:
Ta có:
\(\left\{\begin{matrix} x+2y+3z=4\\ \frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ \frac{6yz+2xy+3xz}{6xyz}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ 2xy+6yz+3xz=0\end{matrix}\right.\)
Do đó:
\((x+2y+3z)^2-2(2xy+6yz+3xz)=4^2-2.0=16\)
\(\Leftrightarrow x^2+4y^2+9z^2=16\)
\(\Leftrightarrow P=16\)
Ta có: \(x+2y+3x=0\Leftrightarrow x=-\left(2y+3z\right)\)
Lại có: \(2xy+6yz+3xz=0\Leftrightarrow x\left(2y+3z\right)+6yz=0\)
\(\Leftrightarrow-\left(2y+3z\right)\left(2y+3z\right)+6yz=0\Leftrightarrow-\left(2y+3z\right)^2+6yz=0\)
\(\Leftrightarrow\left(2y+3z\right)^2-6yz=0\Leftrightarrow4y^2+12yz+9z^2-6yz=0\)
\(\Leftrightarrow4y^2+6yz+9z^2=0\Leftrightarrow\left(2y+\dfrac{3z}{2}\right)^2+\dfrac{27z^2}{4}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+\dfrac{3z}{2}\right)^2=0\\\dfrac{27z^2}{4}=0\end{matrix}\right.\) \(\Rightarrow y=z=0\Rightarrow x=0\)
\(\Rightarrow S=\dfrac{\left(-1\right)^{2019}-1^{2017}+\left(-1\right)^{2015}}{1^{2018}+2.0^{2016}+0^{2014}+2}=\dfrac{-1-1+-1}{1+0+0+2}=\dfrac{-3}{3}=-1\)
x2 + 2y2 + z2 - 2xy - 2y - 4z + 5 = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + ( z2 - 4z + 4 ) = 0
<=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2 = 0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )
Thay ( 1 ) vào A , ta được :
\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)
Vậy A = 2
Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)
\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
\(6=\dfrac{1}{x}+\dfrac{4}{2y}+\dfrac{9}{3z}\ge\dfrac{\left(1+2+3\right)^2}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
theo bài ra ta có
\(\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}=0\Leftrightarrow6yz+3xz+2xy=0\) (1)
\(x+2y+3z=4\Leftrightarrow\left(x+2y+3z\right)^2=16\)
\(\Leftrightarrow x^2+4y^2+9z^2+2\left(6yz+3xz+2xy\right)=16\)(2)
thay (1) vào (2) ta được
\(x^2+4y^2+9z^2=16\)