cho x,y,z$\geq 0 thỏa mãn \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2$
Tìm max abc
05-01-2017 - 19:16
cho x,y,z$\geq 0 thỏa mãn \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=2$
Tìm max abc
05-01-2017 - 19:14
Cho x,y,z>0 thỏa mãn $x^2+y^2+z^2$=1
Tìm min P=$\frac{x}{y^2+z^2}+\frac{y}{x^2+z^2}+\frac{z}{x^2+y^2}$
26-12-2016 - 12:44
Cho x,y,z>0 và $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$
TÌm min P $\frac{x^3}{(y+z)^2}+\frac{y^3}{(x+z)^2}+\frac{z^3}{(y+x)^2}$
26-12-2016 - 12:39
cho x,y,z>0 , x+y+z=1 Tìm max
$\sqrt{\frac{xy}{z+xy}}+\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{xz}{y+xz}}$
13-12-2016 - 15:46
$\left\{\begin{matrix} (x+\sqrt{x^2+4})(y+\sqrt{y^2+1})=2\\ \sqrt{2+xy}+4\sqrt{x}=x+4 \end{matrix}\right.$
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