bài tập cô giao hay quá 10 điểm >-<

thang coppy

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

<=> \(xy+\sqrt{x^2+1}\sqrt{y^2+1}-1=-x\sqrt{x^2+1}-y\sqrt{y^2+1}\)--->Bình phương 2 vế:

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

                                                                                                     \(x^2\left(x^2+1\right)+y^2\left(y^2+1\right)+2xy\sqrt{x^2+1}\sqrt{y^2+1}\)

<=>\(2\left(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\right)=\left(x^2-y^2\right)^2\ge0\)=>\(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\ge0\)

<=>\(1-xy\ge\sqrt{x^2+1}\sqrt{y^2+1}>0\)---> Bình phương 2 vế:

\(1+x^2y^2-2xy\ge\left(x^2+1\right)\left(y^2+1\right)\)<=>\(0\ge\left(x+y\right)^2\ge0\)<=>\(x+y=0\Leftrightarrow x=-y\Rightarrow x^2=y^2\)

--> Thay vào A---> \(A=\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\)

bằng 1 nhá

???

chx chắc đã giòn đâu

?

Ta có : \(S=\frac{20}{x^2+y^2}+\frac{11}{xy}\)

\(=\left(\frac{20}{x^2+y^2}+\frac{10}{xy}\right)+\frac{1}{xy}\)

\(=\left(\frac{20}{x^2+y^2}+\frac{20}{2xy}\right)+\frac{1}{xy}=20.\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}\)

Áp dụng BĐT Svacxo ta có : 

\(20\cdot\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\ge20\cdot\frac{4}{x^2+y^2+2xy}=20\cdot\frac{4}{\left(x+y\right)^2}\ge20\cdot\frac{4}{2^2}=20\)

Mặt khác có : \(0< xy\le\frac{\left(x+y\right)^2}{4}\le\frac{2^2}{4}=1\)

\(\Rightarrow\frac{1}{xy}\ge1\)

Do đó : \(S\ge20+1=21\)

Dấu "=" xảy ra khi \(x=y=1\)

Ez right??

ez game bro then Why u said I am your other account But It true haha

Math with me be hard to win ez to lose

please choa mền xin vài bài đy=)))

Áp dụng bất đăng thức Holder, ta có

\(\Sigma_{cyc} a \sqrt[3]{b^2+c^2} = \Sigma_{cyc} \sqrt[3]{a.a^2.(b^2+c^2)} \le \sqrt[3]{( \Sigma_{cyc} a).(\Sigma_{cyc} a^2).[\Sigma_{cyc} (b^2+c^2)} \le \sqrt[3]{\sqrt{3\Sigma_{cyc} a^2}.(\Sigma_{cyc} a^2).(2\Sigma_{cyc} a^2}) \le 12\)