\(P=\frac{x^3}{1+y}+\frac{y^3}{1+x}\)
x,y dương
xy=1
gtnn
K
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KN
20 tháng 11 2019
a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
KN
20 tháng 11 2019
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)
Ta có:
\(P=\frac{x^3}{1+y}+\frac{y^3}{1+x}=\frac{x^3}{xy+y}+\frac{y^3}{xy+x}=\frac{x^4}{x+1}+\frac{y^4}{y+1}\)
\(=\frac{x^4-1}{x+1}+\frac{y^4-1}{y+1}+\frac{1}{x+1}+\frac{1}{y+1}\)
\(=\left(x^2+1\right)\left(x-1\right)+\left(y^2+1\right)\left(y-1\right)+\frac{1}{x+1}+\frac{1}{y+1}\)
\(=\left(x^2+1\right)\left(x-1\right)+\left(\frac{1}{x^2}+1\right)\left(\frac{1}{x}-1\right)+\frac{1}{x+1}+\frac{1}{\frac{1}{x}+1}\)
\(=\left(x^2+1\right)\left(x-1\right)-\frac{\left(x^2+1\right)\left(x-1\right)}{x^3}+\frac{1}{x+1}+\frac{x}{x+1}\)
\(=\frac{\left(x^2+1\right)\left(x-1\right)\left(x^3-1\right)}{x^3}+\frac{x+1}{x+1}\)
\(\ge\frac{2x\left(x-1\right)^2\left(x^2+x+1\right)}{x^3}+1\)
\(=\frac{2\left(x-1\right)^2.3\sqrt[3]{x^2.x.1}}{x^2}+1=\frac{6\left(x-1\right)^2}{x}+1\)
\(=6\frac{x^2-2x+1}{x}+1=6.\frac{x^2+1}{x}-11\ge12-11=1\)
Dấu "=" xảy ra <=> x = y = 1
Vậy min P = 1 tại x = y = 1.