Áp dụng bđt AM-GM:

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{xz}\)

Nhân theo vế:\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\)

\("="\) khi x=y=z

Khi đó hiển nhiên \(x^3+y^3+z^3=3xyz\)

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

ai giúp mk với! ko hỉu này lắm ạ!

= 2xy(x²-2xy+y²)
= 2xy (x+y)² 
 

Bài 3:

\(\left\{{}\begin{matrix}x+y>=2\sqrt{xy}\\y+z>=2\sqrt{yz}\\x+z>=2\sqrt{xz}\end{matrix}\right.\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)>=8xyz\)

Dấu = xảy ra khi x=y=z

Ta có 4 x 3 y   -   ( - 2 x 3 y )   =   4 x 3 y   +   2 x 3 y   =   6 x 3 y

Chọn đáp án D

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

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

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

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

\(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

2x3y – 2xy3 – 4xy2 – 2xy

= 2xy(x2 - y2 - 2y - 1)

= 2xy[x2 - (y2 + 2y + 1)]

= 2xy[x2 - (y + 1)2 ]

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

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

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

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

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

\(=-4a^2x^3y+6x^3y^2-10x^3y^2z\)