1. Ta có: $x+y+4=0 \Rightarrow x+y=-4$.

Xét: $A=2(x^3+y^3)+3(x^2+y^2)+10xy$.

Ta có: $x^3+y^3=(x+y)^3-3xy(x+y)$ nên: $x^3+y^3=(-4)^3-3xy(-4)=-64+12xy$.

Lại có: $x^2+y^2=(x+y)^2-2xy=16-2xy$.

Thay vào biểu thức $A$:

$A=2(-64+12xy)+3(16-2xy)+10xy$

$=-128+24xy+48-6xy+10xy$

$=-80+28xy$.

Ta có: $(x-y)^2\ge0$

$\Rightarrow (x+y)^2-4xy\ge0$

$\Rightarrow 16-4xy\ge0$

$\Rightarrow xy\le4$.

=> $A=-80+28xy\le-80+28\cdot4=32$.

Dấu “=” xảy ra khi: $x=y=-2$.

Vậy: $\boxed{A_{max}=32}$.

2. Đặt: $t=xy$.

Ta có: $x^4+y^4=(x^2+y^2)^2-2x^2y^2$.

Mà: $x^2+y^2\ge2xy=2t$ nên: $x^4+y^4\ge(2t)^2-2t^2=2t^2$.

Theo giả thiết: $x^4+y^4-7=xy(3-2xy)$

$\Rightarrow x^4+y^4-7=t(3-2t)$.

Do đó: $2t^2-7\le3t-2t^2$

$\Rightarrow 4t^2-3t-7\le0$.

Giải bất phương trình:

$4t^2-3t-7=0$

$\Rightarrow \Delta =(-3)^2-4\cdot4\cdot(-7)=121$

$\Rightarrow \sqrt\Delta=11$.

Suy ra: $t=\dfrac{3\pm11}{8}$

$\Rightarrow t=-1$ hoặc $t=\dfrac74$.

Vì: $4t^2-3t-7\le0$ nên: $-1\le t\le\dfrac74$.

Vậy: $\boxed{M_{min}=-1}$.

\(4M=\left(2x-y-1\right)^2+\left(3y^2+2y+3\right)\)

\(4M=\left(2x-y-1\right)^2+\left[\left(\sqrt{3}y\right)^2+2.\sqrt{3}y.\frac{1}{\sqrt{3}}+\frac{1}{3}\right]+\frac{8}{3}\)

\(4M=\left(2x-y-1\right)^2+\left(\sqrt{3}y+\frac{\sqrt{3}}{3}\right)^2+\frac{8}{3}\)

\(GTNN\left(M\right)=\frac{2}{3}\)

\(khi...y=-\frac{1}{3};x=\frac{1}{3}\)

3. 

P=(x+y)(x^2-xy+y^2)+xy

P=x^2+y^2-xy+xy

P=x^2+y^2

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

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

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

Dấu "=" xảy ra tại x=y=¹/₂

Sửa đề

\(2A=2x^2+2y^2+2xy-2x+2y+2\)

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

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

\(\Rightarrow A_{min}=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(x^2y+xy^2+x+y=xy\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(xy+1\right)=12\left(x+y\right)=2010\)

\(\Rightarrow x+y=\dfrac{2010}{12}\)

\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy=\left(\dfrac{2010}{12}\right)^2-2\cdot11=\dfrac{112137}{4}\)

thêm x²+y²+z²=1 nha

thêm x² + y² + z² = 1 nha

      HT nha vinh