Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)
\(A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\)
\(\ge\frac{4}{x^2+y^2+2xy}+2=\frac{4}{\left(x+y\right)^2}+2=6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\x+y=1\end{cases}}\Rightarrow x=y=\frac{1}{2}\)
tim x, y, z biet
1. \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}\)
2.\(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4x+2}{5}\)va x+y+z=7
bai 1: Tim x biet
\(\hept{\begin{cases}x-y=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\)
bai 2: Tim x, y biet:
x+\(\left(-\frac{31}{12}\right)^2\)=\(\left(\frac{49}{12}\right)^2\)-x=y2
Bai 9: Tim x,y,z biet:
(x-1)2+(x+y)2+(xy-z)2=0
tim gia tri nho nhat cua bieu thuc \(A=\left|1-2x\right|-2009\)
tim 2 so x,y biet \(\frac{x}{y}=\frac{2}{5}\)va x+y=-21
so sanh \(2^{225}vs3^{150}\)
Tìm GTNN của A=\(\frac{1}{x^2+y^2}+\frac{1}{xy}\) biết x,y>0; x+y=1
CHO HAM SO : y=x \(^2\) - 5x +6
a) tim cac gia tri cua x sao cho ve phai cua cong thuc co nghia
b) tinh y ; biet x= -\(\frac{1}{3}\) ; x= 0,5 ; x=0 ; x=1
c) tim x khi biet y = 0
tim GTNN cua B=3-xy biet x+y=1
Tim x,y,z biet:
a, xy=z ; yz=4x ; zx=9x
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Tim cac so Nguyên x va y biet :
a) \(\frac{1}{y}+\frac{x}{4}=\frac{1}{2}\) b)\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
Tim x,y,z biet:
a, \(xy=z;yz=4x;zx=9y\)
b, \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
