Theo bài ra,ta có:

\(xyz=-20\cdot\frac{1}{2}\cdot\frac{1}{5}=-\frac{20}{10}=-2\)

\(\Rightarrow A=-2+\left(-2\right)^2+\left(-2\right)^3+.....+\left(-2\right)^{2019}\)

\(\Rightarrow-2A=\left(-2\right)^2+\left(-2\right)^3+\left(-2\right)^4+....+\left(-2\right)^{2020}\)

\(\Rightarrow-3A=-2^{2020}+2\)

\(\Rightarrow A=\frac{-2^{2020}+2}{-3}\)

x/2=y/3=z/5 và xyz +30

\(5xyz-\dfrac{1}{3}xyz+xyz=\left(5-\dfrac{1}{3}+1\right)xyz=\dfrac{17}{3}xyz\)

\(5xyz-\dfrac{1}{3}xyz+xyz=\dfrac{17}{3}xyz\)

5xyz−1/3xyz+xyz=(5−1/3+1)xyz=17/3xyz

\(=xyz^2\)

đúng không

\(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2+\left(-\frac{1}{4}\right)xyz^2\)

=\(\left(\frac{3}{4}+\frac{1}{2}-\frac{1}{4}\right)xyz^2\)

=\(xyz^2\)

\(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2+\left(-\frac{1}{4}\right)xyz^2\)

=\(xyz^2\left[\frac{3}{4}+\frac{1}{2}+\left(-\frac{1}{4}\right)\right]\)

=\(xyz^2.1\)

= \(xyz^2\)

Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{1}{xy\left(x+y\right)+xyz}+\dfrac{1}{yz\left(y+z\right)+xyz}+\dfrac{1}{zx\left(z+x\right)+xyz}\)

\(\Rightarrow VT\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=\dfrac{1}{x+y+z}.\left(\dfrac{x+y+z}{xyz}\right)=\dfrac{1}{xyz}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)

Với x ; y > 0 , cần c/m : \(x^3+y^3\ge xy\left(x+y\right)\)

Ta có : \(x^3+y^3-xy\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-xy\right)=\left(x+y\right)\left(x-y\right)^2\ge0\)

( điều này luôn đúng với mọi x ; y > 0 )

=> BĐT được c/m

Áp dụng vào bài toán , ta có :

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{xz\left(x+z\right)+xyz}=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{xz\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=\frac{1}{xyz}\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z;x,y,z>0\)

D.\(xyz^2\)

Nhớ tick cho mình nha!