cho \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}},z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}},x\ne0,-1,1\). Hãy biểu diễn z theo y
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ta thấy \(\left(x^2+\frac{1}{x^2}\right)\left(x^2-\frac{1}{x^2}\right)=\left(x^4-\frac{1}{x^4}\right)\)
\(\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right)=\left(x^4+\frac{1}{x^4}\right)+2\)
suy ra \(y=\frac{\left(x^4+\frac{1}{x^4}\right)+2}{\left(x^4-\frac{1}{x^4}\right)}\)
<=> \(y=z+\frac{2}{\left(x^4-\frac{1}{x^4}\right)}\)
<=>\(z=\frac{2}{\left(x^4-\frac{1}{x^4}\right)}-y\)
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a: x-y-z=0
=>x=y+z; y=x-z; z=x-y
\(K=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}=\dfrac{y\cdot\left(-z\right)\cdot x}{xyz}=-1\)
b: Tham khảo:
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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\(\hept{\begin{cases}y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}=\frac{x^4+1}{x^4-1}=a\\z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}=\frac{x^8+1}{x^8-1}\end{cases}}\)
\(\Rightarrow x^4=\frac{y+1}{y-1}\)
Thế vô z được
\(z=\frac{\left(\frac{y+1}{y-1}\right)^2+1}{\left(\frac{y+1}{y-1}\right)-1}=\frac{y^2+1}{2y}\)
Giờ thì thế \(y=\sqrt{2}+\sqrt{3}\)vô đi
Ta có : \(y=\frac{x^2+\frac{1}{x^2}}{x^2-\frac{1}{x^2}}=\frac{x^4+1}{x^4-1}\); \(z=\frac{x^4+\frac{1}{x^4}}{x^4-\frac{1}{x^4}}=\frac{x^8+1}{x^8-1}\)
\(y+\frac{1}{y}=\frac{x^4+1}{x^4-1}+\frac{x^4-1}{x^4+1}=\frac{\left(x^4+1\right)^2+\left(x^4-1\right)^2}{x^8-1}=\frac{2\left(x^8+1\right)}{x^8-1}=2z\)
\(\Rightarrow z=\frac{y+\frac{1}{y}}{2}=\frac{y^2+1}{2y}\)