lớp 12 bạn

8x³-(2x+y).(4x²-2xy+y²)

=\(\left(2x\right)^3-\left(2x+y\right).\left[\left(2x\right)^2-2x.y+y^2\right]\)

= \(\left(2x\right)^3-\left[\left(2x\right)^3+y^3\right]\)

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

= -y³

^{Học tốt !}

\(8x^3-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=8x^3-8x^3-y^3\)

\(=-y^3\)

\(A=4x^2+8x+y^2-4y+20\)

\(A=\left(4x^2+8x\right)+\left(y^2-4y\right)+20\)

\(A=4\left(x^2+2x+1\right)+\left(y^2-4y+4\right)-4-4+20\)

\(A=4\left(x+1\right)^2+\left(y-2\right)^2+12\ge12\forall x,y\)

Do \(4\left(x+1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\)

Dấu "=" Xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy Min A=12 <=>\(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

a)  ... = (x^2 -2xy + y^2)+(x^2 -2x+1)+2014=(x-y)^2 + (x-1)^2 +2014 >= 2014 

Đăngt thức xay ra khi x=y=1

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+x^2+6x+9+1978\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x+3\right)^2+1978\)

\(=\left(x-y+1\right)^2+\left(x+3\right)^2+1978\ge1978\)

\(A_{min}=1978\) khi \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)