mk chịu ?_?

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

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

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

Dấu "=" xảy ra tại x=y=¹/₂

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We have:

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y+3\right)^2=-y^2+1\)

\(\Rightarrow\left(x+y+3\right)^2\le1\)

\(\Rightarrow-1\le x+y+3\le1\)

\(\Rightarrow2015\le x+y+2019\le2017\)

Sign '=' happen when \(x=-4;x=-2;y=0\)

A nguyên =>\(1-\frac{2}{x-2}\) nguyên

=>\(\frac{2}{x-2}\)nguyên 

=>x-2 thuộc Ư(2) thuộc (1;2;-1;-2;0)

=>x thuộc (3;4;1;0;2)

Kết luận:....................

#Châu's ngốc

Ta có: \(\left(x-y\right)^2\ge0\)

\(\Rightarrow x^2-2xy+y^2\ge0\)

\(\Rightarrow x^2+2xy+y^2\ge4xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

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

Ta có vì : x,y > 0

và \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Từ đề bài ta có:

\(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\frac{x+y}{xy}.\left(x+y\right).xy\ge\frac{4}{x+y}.xy\left(x+y\right)\)

Áp dụng đẳng thức Cô-si:

\(\Leftrightarrow x^2+2xy+y^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)

Vậy....

đpcm.

\(-10x^3y\left(\frac{2}{5}x^2y+\frac{3}{10}xy^2\right)+3x^4y^3\)

\(=-4x^5y^2-3x^4y^3+3x^4y^3=-4x^5y^2\)

\(-10x^3y\left(\frac{2}{5}x^2y+\frac{3}{10}xy^2\right)+3x^4y^3\)

\(=\left[\left(-10x^2\right)\left(y\right)\right].\left[\left(\frac{2}{5}x^2\right)\left(y\right)+\left(\frac{3}{10}x\right)\left(y^2\right)\right]+3x^4y^3\)

\(=-4x^5y^2-3x^4y^3+3x^4y^3\)

\(=-4x^5y^2\)

dit me may