Cho x , y , z thỏa mãn : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\) và \(\frac{2}{xy}-\frac{1}{z^2}=4\) Tính D = \(\left(x+2y+z\right)^{2018}\)
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21 tháng 8 2019
+ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\frac{1}{z}=2-\frac{1}{x}-\frac{1}{y}\)
\(\Rightarrow\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\)
+ \(\frac{2}{xy}-\frac{1}{z^2}=4\Rightarrow\frac{2}{xy}-\left(2-\frac{1}{x}-\frac{1}{y}\right)^2=4\)
\(\Rightarrow\frac{2}{xy}-\left(4+\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+\frac{2}{xy}\right)=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+8=0\)
\(\Rightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\) \(\Rightarrow\left\{{}\begin{matrix}\left(\frac{1}{x}-2\right)^2=0\\\left(\frac{1}{y}-2\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=2\\\frac{1}{y}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\\frac{1}{z}=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow P=\left(\frac{1}{2}+1-\frac{1}{2}\right)^{2018}=1\)
G
9 tháng 3 2021
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)
\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\)
\(\Leftrightarrow z=\frac{-1}{2}\)
\(x=y=\frac{1}{2}\)
\(\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1\)
NC
9 tháng 3 2021
Ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)
\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\\\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=-\frac{1}{z}\\\frac{1}{y}=-\frac{1}{z}\end{cases}}}\)
\(\Leftrightarrow x=y=-z\)
Thay vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)ta được :
\(x=y=\frac{1}{2};z=-\frac{1}{2}\)
\(\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1\)
7 tháng 2 2019
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
4 tháng 11 2017
vì x+y+z=1nên
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\)\(\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)\(=3+\left(\frac{x}{y}+\frac{y}{z}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)=\(3+\frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{x^2+z^2}{xz}\)
nen \(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{xz}{x^2+z^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) =\(\left(\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}\right)+\left(\frac{yz}{y^2+z^2}+\frac{y^2+z^2}{4yz}\right)+\left(\frac{xz}{x^2+z^2}+\frac{x^2+z^2}{xz}\right)+\frac{3}{4}\)
\(\ge2.\frac{1}{2}+\frac{2.1}{2}+\frac{2.1}{2}+\frac{3}{4}=\frac{15}{4}\)(dpcm)
dau = xay ra khi x=y=z=1/3