Ta có: xyz + x2y2z2 + x3y3z3 + ….. + x10y10z10

      = xyz + (xyz)2 + (xyz)3 + ….. + (xyz)10

Với x = 1; y = -1; z = - 1 ta có: xyz = 1.(-1).(-1) = 1

Thay vào đa thức: 1 + 12 + 13 + … + 110 = 10

a, 3y(x2-xy)-7x2(y+xy)

= 6xy - 3xy² - 14xy - 14x²y

=-8xy-3xy²-14x²y

Bậc: 2

3x2y2z2 + x2y2z2 = (3 + 1) x2y2z2 = 4 x2y2z2

a. x2 + 5x2 + (-3x2) = (1 + 5 – 3)x2 = 3x2

b. 5xy2 + 1/2 xy2 + 1/4 xy2 + (-1/2 )xy2 = (5 + 1/2 + 1/4 - 1/2 )xy2 = 21/4 xy2

c. 3x2y2z2 + x2y2z2 = (3 + 1) x2y2z2 = 4 x2y2z2

a. x2 + 5x2 + (-3x2) = (1 + 5 – 3)x2 = 3x2

b. 5xy2 + 1/2 xy2 + 1/4 xy2 + (-1/2 )xy2 = (5 + 1/2 + 1/4 - 1/2 )xy2 = 21/4 xy2

c. 3x2y2z2 + x2y2z2 = (3 + 1) x2y2z2 = 4 x2y2z2

4/xyz vì xyz cùng mẫu

k nha

\(\frac{1}{xyz}+\frac{1}{xyz}+\frac{2}{xyz}=\frac{1+1+2}{xyz}=\frac{4}{xyz}\)

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 \(x=y=z\)

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

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

5xyz−1/3xyz+xyz=(5−1/3+1)xyz=17/3xyz

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 \(\Leftrightarrow x=y=z;x,y,z>0\)

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