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

a) ĐKXĐ: \(\hept{\begin{cases}x-2\ne0\\x\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

b)Ta có: P = \(\frac{x^2}{x-2}\left(\frac{x^2+4}{x}-4\right)+3\)

P = \(\frac{x^2}{x-2}\left(\frac{x^2+4-4x}{x}\right)+3\)

P = \(\frac{x^2}{x-2}\cdot\frac{\left(x-2\right)^2}{x}+3\)

P = \(\left(x-2\right).x+3\)

P = \(x^2-2x+3\)

c) Ta có: P = x² - 2x + 3

P = (x² - 2x + 1) + 2

P = (x - 1)² + 2 \(\ge\)2 \(\forall\)x

Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1

Vậy x = 1 thì P đạt GTNN là 2

a) Phân thức được xác định khi \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

ĐKXĐ: \(x\ne2;x\ne0\)

b) \(P=\frac{x^2}{x-2}\left(\frac{x^2+4}{x}-4\right)+3\)

\(P=\frac{x^4-4x^3+7x^2-6x}{x^2-2x}\)

\(P=\frac{x^3-4x^2+7x-6}{x-2}\)

\(P=\frac{\left(x-2\right)\left(x^2-2x+3\right)}{x-2}\)

\(P=x^2-2x+3\)

c) \(P=x^2-2x+3\)

\(P=x^2-2x+1+2\)

\(P=\left(x-1\right)^2+2\ge2\) vì \(\left(x-1\right)^2\ge0,\forall x\inℝ\)

\(\Rightarrow Min_P=2\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy: \(Min_p=2\Leftrightarrow x=1\)

\(P=\frac{n^3+2n-1}{n^3+2n^2+2n+1}\)

\(=\frac{n^3+2n-1}{\left(n^3+1\right)+\left(2n^2+2n\right)}\)

\(=\frac{n^3+2n-1}{\left(n+1\right)\left(n^2-n+1\right)+2n\left(n+1\right)}\)

\(=\frac{n^3+2n-1}{\left(n+1\right)\left(n^2+n+1\right)}\)

Để phân thức xác định thì \(n+1\ne0\Rightarrow n\ne1\)

(vì \(n^2+n+1=\left(n+\frac{1}{2}\right)^2+\frac{3}{4}>0\))

3x(x-3)+5(x-3)=0

\(\Leftrightarrow\)(x-3)(3x+5)=0

\(\Leftrightarrow\)x-3=0 hoặc 3x+5=0

\(\Leftrightarrow\)x=0+3 hoặc 3x=-5

\(\Leftrightarrow\)x=3 hoặc x=-5/3

Vậy x=3; x=-5/3

\(3x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\)  \(\left(x-3\right)\left(3x+5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=0\\3x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{3}\end{cases}}\)

Vậy \(S=\left\{3;\frac{-5}{3}\right\}\)

Bước 1) Nhóm nhân tử chung

Bước 2) Vì VP là 0 => 2TH bằng 0

Bước 3) Tìm x

Bước 4) KL

Cảm ơn! Mong là mình làm đúng.

Ta có:\(\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)

\(\ge\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{b+c+d+a}+\frac{d}{d+a+b+c}=1\)

và  \(\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)

\(\le\frac{a}{a+c}+\frac{b}{b+d}+\frac{c}{c+a}+\frac{d}{d+b}\)

\(=1+1=2\)

Vậy \(1\le\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\le2\)(đpcm)

dit me may