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26 tháng 2 2018
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
BT
8 tháng 2 2017
A=\(\left[\frac{x\left(x-y\right)}{y\left(x+y\right)}+\frac{\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\right]:\left[\frac{y^2}{x\left(x-y\right)\left(x+y\right)}+\frac{1}{x+y}\right]\frac{ }{ }\)
=\(\left[\frac{x^2\left(x-y\right)+y\left(x-y\right)\left(x+y\right)}{xy\left(x+y\right)}\right]:\left[\frac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\right]\)=\(\frac{\left(x-y\right)\left(x^2+y^2+xy\right)}{xy\left(x+y\right)}.\frac{x\left(x-y\right)\left(x+y\right)}{y^2+x\left(x-y\right)}\)
=\(\frac{\left(x-y\right)^2\left(x^2+y^2+xy\right)}{y\left(x^2+y^2-xy\right)}\)=\(\frac{\left(x-y\right)^2\left(x^2+xy+\frac{y^2}{4}+\frac{3y^2}{4}\right)}{y\left(x^2-xy+\frac{y^2}{4}+\frac{3y^2}{4}\right)}\)=\(\frac{\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{y.\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}\)
Ta nhận thấy các số trong ngoặc đều dương.
=> Để A>0 thì y>0
Vậy để A>0 thì y>0 và với mọi x
LT
0
8 tháng 2 2020
\(x^2=yz\Rightarrow\frac{x}{y}=\frac{z}{x}\left(1\right)\)
\(y^2=xz\Rightarrow\frac{x}{y}=\frac{y}{z}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
\(\Rightarrow x=y=z\)
Thay y, z bằng x \(\Rightarrow M=\frac{3.x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
TT
1
26 tháng 12 2016
=\(\left(\frac{1}{x\left(x-y\right)}-\frac{3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(\left(\frac{y\left(x+y\right)+x^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy+y^2-3y^2-y\left(x-y\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\) \(\left(\frac{x^2+xy+y^2}{x+y}\right)\)
=\(\left(\frac{x^2+xy-2y^2-xy+y^2}{x\left(x-y\right)}\right)\left(\frac{1}{x+y}\right)\)
=\(\frac{x^2-y^2}{x\left(x-y\right)\left(x+y\right)}\)=\(\frac{\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}\) =\(\frac{1}{x}\)
25 tháng 12 2021
\(x+y+z=1\\ \Rightarrow\left\{{}\begin{matrix}x=1-y-z\\y=1-x-z\\z=1-x-y\end{matrix}\right.\)
\(S=\dfrac{\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)}{\left(1-x\right)^2\left(1-y\right)^2\left(1-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left(xy+1-x-y\right)\left(yz+1-y-z\right)\left(zx+1-x-z\right)}{\left(x+y+z-x\right)^2\left(x+y+z-y\right)^2\left(x+y+z-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left[\left(xy-x\right)-\left(y-1\right)\right]\left[\left(yz-y\right)-\left(z-1\right)\right]\left[\left(zx-x\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left[x\left(y-1\right)-\left(y-1\right)\right]\left[y\left(z-1\right)-\left(z-1\right)\right]\left[x\left(z-1\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)\left(y-1\right)\left(y-1\right)\left(z-1\right)\left(x-1\right)\left(z-1\right)}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)^2\left(y-1\right)^2\left(z-1\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-x-y-z\right)^2\left(y-x-y-z\right)^2\left(z-x-y-z\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(-y-z\right)^2\left(-x-z\right)^2\left(-x-y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=1\)