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

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

b, ta có : \(x+y=1=>2x+2y=2\)

\(B=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{4}{4x^2+4y^2}+\dfrac{6}{8xy}\)\(\ge\dfrac{\left(2+\sqrt{6}\right)^2}{\left(2x+2y\right)^2}\)

\(=\dfrac{\left(2+\sqrt{6}\right)^2}{2^2}=\dfrac{5+2\sqrt{6}}{2}\)=>\(B\ge\dfrac{5+2\sqrt{6}}{2}\)

=>\(MinB=\dfrac{5+2\sqrt{6}}{2}\)

Ai giúp mik vs

Huhu ai giúp vs

bạn sửa lại là 9-2t^2 nhé , mình đánh nhầm ^^

chuẩn nhé !

Hệ \(\hept{\begin{cases}x+y\le2\\x^2+y^2+xy=3\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=2-a\left(a\ge0\right)\\x^2+y^2+xy=3\end{cases}}}\)

Do đó \(\hept{\begin{cases}x+y=2-a\\xy=\left(2-a\right)^2-3\end{cases},\Delta=S^2-4P\ge0\Rightarrow0\le a\le4}\)

\(T=x^2+y^2+xy-2xy=9-2\left(2-a\right)^2\)

minT=1 khi x=1; y=1 hoặc x=-1; y=-1

maxT=9 khi \(\orbr{\begin{cases}x=\sqrt{3};y=-\sqrt{3}\\x=-\sqrt{3};y=\sqrt{3}\end{cases}}\)

ĐK: x khác 0

Từ\(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)

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

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

Do VT > 0\(\Rightarrow6+xy\ge0\Rightarrow xy\ge6\)
Có A = 2016 + xy > 2016 + 6 = 2022

tth : Viết nhầm :V
Đoạn cuối \(6+xy\ge0\Rightarrow xy\ge-6\)

Có A = 2016 + xy > 2016 - 6 = 2010 !!!

Được rồi chứ gì -.- 

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{1}{x}=0\\x+\frac{y}{2}=0\end{cases}}\)

             \(\Leftrightarrow\hept{\begin{cases}x^2=1\\x=-\frac{y}{2}\end{cases}}\)

             \(\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\left(h\right)\hept{\begin{cases}x=-1\\y=2\end{cases}}\)OK ???

\(P=\sqrt{x+1}+\sqrt{y+1}\ge\sqrt{x+1+y+1}=\sqrt{x+y+2}=\sqrt{101}\)

GTNN\(P=\sqrt{101}\)

\(P=\sqrt{x+1}+\sqrt{y+1}\)

\(=>\left(\sqrt{x+1}+\sqrt{y+1}\right)^2\le2\left(x+1+y+1\right)=2.101=202\)

GTLN \(P=202\)