Bài này mình đánh bị lỗi nha

Lỗi đề ak bn??

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

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

\(=1+\frac{1-x^2-y^2}{x^2y^2}=1+\frac{\left(x+y\right)^2-x^2-y^2}{x^2y^2}=1+\frac{2}{xy}\)

Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow1+\frac{2}{xy}\ge1+\frac{2}{\frac{1}{4}}=9\)

\("="\Leftrightarrow x=y=\frac{1}{2}\)

Áp dụng BĐT AM-GM ta có:

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

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{\left(1+4\right)^2}{2}=\frac{5^2}{2}=\frac{25}{2}\)

Xảy ra khi \(x=y=\frac{1}{2}\)

\(1=2\left(\dfrac{9}{x}+\dfrac{1}{y}\right)\ge2.\dfrac{\left(3+1\right)^2}{x+y}=\dfrac{32}{x+y}\)

\(\Rightarrow x+y\ge32\)

\(A_{min}=32\) khi \(\left(x;y\right)=\left(24;8\right)\)