a: \(x^3+y^3+xy\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+xy\)

\(=1-3xy+xy=-2xy+1\)

b: \(x^3-y^3-xy\)

\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\)

\(=1+3xy-xy=2xy+1\)

ez

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

\(\Leftrightarrow2\left(x^2+y^2-xy\right)=2\left(x+y-1\right)\)

\(\Leftrightarrow2x^2+2y^2-2xy=2x+2y-2\)

\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y+2=0\)

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

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

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

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}1-1=0\left(tm\right)\\x=1\\y=1\end{cases}}}\)

\(\Rightarrow M=\left(1+1\right)^{10}=2^{10}=1024\)

C=720

Lời giải:

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

\(=(x-y)(x^2+xy+y^2)=x^3-y^3=10^3-(-1)^3=1000-(-1)=1001\)

\(C=x^4+10x^3+10x^2+10\)

\(=x^4+9x^3+x^3+9x^2+x^2+10\)

\(=x^3(x+9)+x^2(x+9)+x^2+10\)

\(=(x+9)(x^3+x^2)+x^2+10\)

\(=(-9+9)[(-9)^3+(-9)^2]+(-9)^2+10\)

\(=0+(-9)^2+10=91\)

Thay $x=-1$ vào biểu thức:

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

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

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

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

\(=(x+1)(y+1)=(-1+1)(y+1)=0\)

Áp dụng BĐT : ( a + b + c )² \(\ge\)3 ( ab + bc + ac )

Ta có : \(\frac{\left(x+y+1\right)^2}{xy+y+x}\ge\frac{3\left(xy+y+x\right)}{xy+y+x}=3\)

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

ta có : \(A+\frac{1}{A}=\frac{8A}{9}+\frac{A}{9}+\frac{1}{A}\ge\frac{8.3}{9}+2\sqrt{\frac{A}{9}.\frac{1}{A}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

=> \(a^2+b^2+c^2\ge ab+bc+ac\)=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

Áp dụng ta được

\(\left(x+y+1\right)^2\ge3\left(x+y+xy\right)\)=> \(\frac{\left(x+y+1\right)^2}{xy+y+x}\ge3\)

Đặt \(\frac{\left(x+y+1\right)^2}{x+y+xy}=t\)(\(t\ge3\))

Khi đó

\(VT=t+\frac{1}{t}=\left(\frac{t}{9}+\frac{1}{t}\right)+\frac{8}{9}t\ge\frac{2}{3}+\frac{8}{9}.3=\frac{10}{3}\)

Dấu bằng xảy ra khi \(\hept{\begin{cases}t=3\\x=y=1\end{cases}}\)=> x=y=1

Lưu ý 

Nhiều người sẽ nhầm \(VT\ge2\)

Khi đó dấu bằng \(\left(x+y+1\right)^2=xy+x+y\)không xảy ra 

a: \(\frac{y}{x^2-xy}+\frac{x}{y^2-xy}\)

\(=\frac{y}{x\left(x-y\right)}-\frac{x}{y\left(x-y\right)}\)

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

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

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

\(=\frac{x-y-2x-\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=\frac{-x-y-x-y}{\left(x+y\right)\left(x-y\right)}=\frac{-2x-2y}{\left(x+y\right)\left(x-y\right)}\)

\(=\frac{-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{-2}{x-y}\)

Từ \(x^2+y=y^2+x\Leftrightarrow x^2-y^2-x+y=0\)

<=>(x-y)(x+y)-(x-y)=0

<=>(x-y)(x+y-1)=0

Vì x khác y => x+y-1=0

<=>x+y=1

<=> (x+y)²=1

<=> x²+y²=1-2xy

Thay vào A ta được: \(A=\frac{1-2xy+xy}{xy-1}=\frac{1-xy}{xy-1}=\frac{-\left(xy-1\right)}{xy-1}=-1\)

x#y sao lại suy ra x+y-1=0

Ta có :\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow3x^2+3y^2=10xy\)

\(\Rightarrow M^2=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\frac{10xy-6xy}{10xy+6xy}=\frac{4xy}{16xy}=\frac{1}{4}\)

Vậy M=\(\frac{1}{4}\)