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Lời giải cho bài của bạn ở đây nhé! http://olm.vn/hoi-dap/question/479780.html
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\(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{3}{z}=0\)
=>\(\dfrac{yz+2xz+3xy}{xyz}=0\)
=>yz+2xz+3xy=0
=>\(xy+\dfrac{2}{3}xz+\dfrac{1}{3}yz=0\)
\(x+\dfrac{y}{2}+\dfrac{z}{3}=1\)
=>\(\left(x+\dfrac{y}{2}+\dfrac{z}{3}\right)^2=1\)
=>\(x^2+\dfrac{y^2}{4}+\dfrac{z^2}{9}+2\left(x\cdot\dfrac{y}{2}+x\cdot\dfrac{z}{3}+\dfrac{y}{2}\cdot\dfrac{z}{3}\right)=1\)
=>\(A+2\left(\dfrac{xy}{2}+\dfrac{xz}{3}+\dfrac{yz}{6}\right)=1\)
=>A+xy+2/3xz+1/3yz=1
=>A=1
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Bạn tham khảo ở đây.
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\(x+y+z=9\Leftrightarrow\left(x+y+z\right)^2=81\\ \Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=81\\ \Leftrightarrow xy+yz+xz=\dfrac{81-27}{2}=27\\ \Leftrightarrow x^2+y^2+z^2=xy+yz+xz\\ \Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z=\dfrac{9}{3}=3\left(x+y+z=9\right)\)
\(\Leftrightarrow\left(x-4\right)^{2018}+\left(y-4\right)^{2019}+\left(z-4\right)^{2020}\\ =\left(-1\right)^{2018}+\left(-1\right)^{2019}+\left(-1\right)^{2020}=1-1+1=1\)
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Ta có: \(x+y+z=0\)
\(\Leftrightarrow\) \(\left(x+y+z\right)^2=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2=0\) (vì xy + yz + xz =0)
\(\Leftrightarrow\)\(x=y=z=0\)
Vậy \(S=\left(0-1\right)^{1999}+0^{2003}+\left(0+1\right)^{2006}=0\)
\(x+y+z=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)\(=0\)
\(\Rightarrow2xy+2yz+2xz=-9\)
\(\Rightarrow xy+yz+xz=-\frac{9}{2}\)
\(\Rightarrow x^2y^2+y^2z^2+x^2z^2+2xy^2z+2xyz^2+2x^2yz=\left(-\frac{9}{2}\right)^2=\frac{81}{4}\)\(\)
\(\Rightarrow x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)=\frac{81}{4}\)
\(\Rightarrow x^2y^2+y^2z^2+x^2z^2=\frac{81}{4}\)
\(\left(x^2+y^2+z^2\right)^2=9^2=81\)
\(\Rightarrow P=x^4+y^4+z^4=81-2\left(x^2y^2+y^2z^2+x^2z^2\right)=81-2.\frac{81}{4}=\frac{81}{2}\)