Question

 

Answer 1

lỗi 

Answer 2

lỗi rùi

Answer 3

\(A=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)=\left(1-\dfrac{1}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1+\dfrac{1}{y}\right)\)

\(=\left(1-1-\dfrac{y}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-1-\dfrac{x}{y}\right)\left(1+\dfrac{1}{y}\right)\)

\(=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\dfrac{1}{y}+\dfrac{1}{x}+1+\dfrac{1}{xy}=\dfrac{x+y+1}{xy}+1=\dfrac{2}{xy}+1\)

áp dụng bđt cosi \(x+y\ge2\sqrt{xy}< =>1\ge2\sqrt{xy}< =>xy\le\dfrac{1}{4}< =>\dfrac{2}{xy}\ge8\)

<=> A >= 8+1=9 

dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)