Cho x,y,z>0 và xy+yz+zx=1
Tính giá trị bt:
\(P=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Cho x,y,z thỏa mãn xy+yz+xz = 1.Tính
\(S=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
cho x,y,z thỏa mãn \(x+y+z\le\dfrac{3}{2}\) . tìm GTNN của \(P=\dfrac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\dfrac{y\left(xz+1\right)^2}{y^2\left(xy+1\right)}+\dfrac{z\left(xy+1\right)^2}{x^2\left(yz+1\right)}\)
Cho x, y, z dương thỏa mãn xyz=1. Tìm GTLN của \(\dfrac{1}{\sqrt{\left(x+y\right)^2+\left(x+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(y+z\right)^2+\left(y+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(z+x\right)^2+\left(z+1\right)^2+4}}\)
Cho x,y,z>0; x+y+z=1
Tính \(Q=\sqrt{\dfrac{\left(x+yz\right)\left(y+xz\right)}{xy+z}}+\sqrt{\dfrac{\left(y+xz\right)\left(z+xy\right)}{x+yz}}+\sqrt{\dfrac{\left(x+yz\right)\left(z+xy\right)}{y+xz}}\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Giải hệ phương trình sau : \(\left\{{}\begin{matrix}\sqrt{2x+1}+\sqrt{2y+1}=\dfrac{\left(x-y\right)^2}{2}\\\left(3x+2y\right)\left(y+1\right)=4-x^2\end{matrix}\right.\)
Cho 3 số dương x;y;z thỏa mãn x+y+z=6. CMR: \(x^2+y^2+z^2-xy-yz-xz+xyz\ge8\)
Tìm GTLN của biếu thức
P= x\(\sqrt{\dfrac{\left(1+y^2\right).\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right).\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right).\left(1+y^2\right)}{1+z^2}}\)
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)