em viết nhầm đề nha.M = \(\frac{y}{\sqrt{xy}-x}+\frac{x}{\sqrt{xy}+y}-\frac{x+y}{\sqrt{xy}}\)mới đúng

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

\(M=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

\(\ge\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{\left(x+y\right)^2}{2}}=\frac{4}{1}+\frac{1}{\frac{1}{2}}=6\)

mình không hiểu quá trình biến đổi của bạn alibaba sau dấu lớn hơn hoạc bằng

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

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

\(\ge\frac{4}{x^2+y^2+2xy}+2=\frac{4}{\left(x+y\right)^2}+2=6\)

Dấu "=" xảy ra khio x=y=1/2

1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

b/

\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)

\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)

\(=16+8+20=44\)

\(\Rightarrow B\ge11\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

2/

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

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{25}{2}\)

Dấu = xảy ra khi \(x=y=\frac{1}{2}\)

???????

\(A=\frac{1}{x^2+y^2}+\frac{5}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{9}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{9}{2\left(\frac{x+y}{2}\right)^2}\)

nên  \(A\ge4+9.2=22\)

Dấu bằng xảy ra khi và chỉ khi  \(x=y=\frac{1}{2}\)

chịu thua vô điều kiện xin lỗi nha : v

muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v

\(A=\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]:\frac{\sqrt{x^3}+y.\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(\Leftrightarrow A=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{x+y}{xy}\right]:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{2\sqrt{xy}+x+y}{xy}:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{xy}\left(x+y\right)}{xy\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Leftrightarrow A=\frac{\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

sai sót chỗ nào chỉ cho mk nhé. ý kia chốc nx làm nốt