tim x, y nguyen duong thoa man xy+ x- y = 4
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a)Tim cap (x,y) nguyen duong thoa man xy=3(y-x)
b)cho 2 so x,y >0 thoa man x+y = 1
Tim GTNN cua M=(x^2+1/y^2)(y^2+1/x^2)
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Tim so nguyen duong x,y thoa man dang thuc
x2+xy-2=0
cho x, y nguyen duong thoa man xy-5x+2y=30. khi do tong cac gia tri cua x tim duoc la?
tim cac so nguyen duong x,y,z thoa man x^2+y^3+z^4
tim so nguyen duong x,y thoa man : x^2=1!+2!+3!+....+y!
Tim cac so nguyen duong x;y;z thoa man x!+y!=10.z+9
tim cac so huu ti x y thoa man x+y va 1/x +1/y nguyen duong
Cho x,y,z nguyen duong thoa man x+y-z+1=0
Tim GTLN cua \(P=\frac{x^3y^3}{\left(x+yz\right)\left(y+xz\right)\left(z+xy\right)^2}\)
Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)
\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)
\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)
Áp dụng BĐT Cosi ta có:
\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)
\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)
tim cap so nguyen (x,y) thoa man: x+y+xy=2