Quay lại đáp án

có P = 1+x + $\frac{x^{2}}{1-x}$+1+y + $\frac{y^{2}}{1-y}$ + $\frac{1}{x+y}$ -2 = $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ -2 có ((1-x)+(1-y)+(x+y)) ( $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y})$ $\geq$ 9 $\Rightarrow$ $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ $\geq$ $\frac{9}{2}$ $\Rightarrow$ P $\geq$ $\frac{5}{2}$ dấu "=" $\Leftrightarrow$ x=y= $\frac{1}{3}$

1

có P = 1+x + $\frac{x^{2}}{1-x}$+1+y + $\frac{y^{2}}{1-y}$ + $\frac{1}{x+y}$ -2 = $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ -2 có ((1-x)+(1-y)+(x+y)) ( $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ $\geq$ 9 $\Rightarrow$ $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ $\geq$ $\frac{9}{2}$ $\Rightarrow$ P $\geq$ $\frac{5}{2}$ dấu "=" $\Leftrightarrow$ x=y= $\frac{1}{3}$

có P = 1+x + $\frac{x^{2}}{1-x}$+1+y + $\frac{y^{2}}{1-y}$ + $\frac{1}{x+y}$ -2 = $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y}$ -2 có ((1-x)+(1-y)+(x+y)) ( $\frac{1}{1-x}$ + $\frac{1}{1-y}$ + $\frac{1}{x+y})\geq$9$\Rightarrow\frac{1}{1-x}$+$\frac{1}{1-y}$+$\frac{1}{x+y}\geq\frac{9}{2}\Rightarrow$P$\geq\frac{5}{2}$dấu "="$\Leftrightarrow$x=y=$\frac{1}{3}$