\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Vì \(\left(x+y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+1\right)^2\ge0\)

\(\Rightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}=-1\)

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

mk ko vt lại đề 

=> (4x^2+8xy+4y^2)+(x^2-2x+1)+(y^2+2y+1)=0

=>(2x+2y)^2+(x-1)^2+(y+1)^2=0

...... phần này bn tự làm đc

=>x=1,y=-1

thay vào là dc

Ta có : \(5x^2+5y^2+8xy-2x+2y+2=0\)

=> \(\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)=0\)

=> \(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Ta có \(\left(2x+2y\right)^2\ge0\forall x,y\)   ,   \(\left(x-1\right)^2\ge0\forall x\)   ,   \(\left(y+1\right)^2\ge0\forall x\)

=> \(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

=> \(\hept{\begin{cases}x+y=0\\x-1=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}}\)

Thay vào M ta có:

\(M=0^{2016}+\left(1-2\right)^{2018}+\left(-1+1\right)^{2019}=1\)

By Titu's Lemma we easy have:

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

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

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

\(=\frac{17}{4}\)

Mk xin b2 nha!

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

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

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

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

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

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))

\(\Leftrightarrow x=4y\)

Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)

\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1