Violympic toán 8
Các câu hỏi tương tự
Cho: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\) và x+y+z=xyz (x, y, z khác 0). CM: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=2\)
Cho x, y, z>0 và \(x+y+z\le1\). CM: \(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge10\)
cho x,y,x>0
cm: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}< =\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Cho x,y,z>0 . Cm : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{16}{x+y+z}\)
cho x,y,z thỏa mãn \(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=1\)
Cm: \(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}=0\)
Cho x,y,z # 0 và\(\dfrac{x-y-z}{x}=\dfrac{y-x-z}{y}=\dfrac{-x-y+z}{z}\)
Tính A =\(\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
Cho các số dương x, y, z thỏa mãn: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). CM: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\)
Cho các số dương x, y, z thỏa mãn: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). CM: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\)
cho 3 so thoa man \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
CM: \(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}=0\)