tim GTLN \(\frac{x^2+xy+y^2}{x^2-xy+y^2}\)
Tim GTLN : E=\(\frac{x^2+xy+y^2}{x^2-xy+y^2}\)voi x,y>0
Tim GTLN : M=\(\frac{x}{\left(x+1995\right)^2}\)voi x>0
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cho x,y>0 và xy=1. Tim GTLN A=x^2+3x+y^2+3y+\(\frac{9}{x^2+y^2+1}\)
tim dieu kien cua x va y de A khong am
\(A=\left(\frac{x^2-xy}{y=xy}+\frac{x^2-y^2}{x^2+xy}\right):\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x-y}\right)\)
Tìm GTLN
\(A=\frac{x^2+xy+y^2}{x^2-xy+y^2}\) \(x;y>0\)
sửa \(x,y\ge0\)
ta có \(A=\frac{3x^2+3xy+3y^2}{3\left(x^2-xy+y^2\right)}=\frac{x^2-xy+y^2+2x^2+4xy+2y^2}{3\left(x^2-xy+y^2\right)}=\frac{1}{3}+\frac{2\left(x+y\right)^2}{3\left(x^2-xy+y^2\right)}\)
ta có \(\frac{2\left(x+y\right)^2}{3\left(x^2-xy+y^2\right)}\ge0\) => \(A\ge\frac{1}{3}\)
dấu = xảy ra <=> x=-y
Cho x,y,z >0 tm xy+yz+zx=xyz. Tìm GTLN của:
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
tim cap so nguyen x,y
x+y+xy=2
b, tim gtln cua bieu thuc
q=\(\frac{27-2x}{12-x}\)(x nguyen)
(Hòa Bình)
Tìm GTLN và GTNN của biểu thức \(P=\frac{x^2-xy+y^2}{x^2+xy+y^2}\).
Cho x,y,z > 0. Tìm GTLN của: \(A=\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{xz}}{y+2\sqrt{xz}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}\)
Cho x, y, z > 0. Tìm GTLN của \(A=\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\)
\(A=\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\)
\(2A=\frac{z+2\sqrt{xy}}{z+2\sqrt{xy}}-\frac{z}{z+2\sqrt{xy}}+\frac{x+2\sqrt{yz}}{x+2\sqrt{yz}}-\frac{x}{x+2\sqrt{yz}}+\frac{y+2\sqrt{zx}}{y+2\sqrt{zx}}-\frac{y}{y+2\sqrt{zx}}\)
\(=3-\left(\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{zx}}+\frac{z}{z+2\sqrt{xy}}\right)\le3-\left(\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\right)\)
\(=3-\frac{x+y+z}{x+y+z}=3-1=2\)\(\Leftrightarrow\)\(A\le\frac{2}{2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)
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