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

\(\Leftrightarrow\hept{\begin{cases}x+2y=0\\y-1=0\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=0\\y=1\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\\x-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\\z=-2\end{cases}}\)

Do đó: \(x+2y+3\text{z}=-2+2-2.3=-6\)

Vậy: \(M=-6\)

3x²y²z² = x³y³ y³z³ z³x³ 
(3x²y²z²) / (x³y³ y³z³ z³x³) = 1
3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1
(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3
(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3
z²/(xy) x/(yz) y²/(zx) = 1/3
Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3

Ta có: H = x³ + x²y - xy² - y³ + x² - y² + 2x + 2y + 4 

= x²(x + y) - y²(x + y) + (x² - y²) + 2(x + y + 2)

= (x + y)(x² - y²) + (x² - y²) + 2(x + y + 1 + 1)

= (x + y + 1)(x² - y²) + 2(0 + 1)

= 0(x² - y²) + 2.1

= 2

Vậy H = 2

Chúc bn học tốt!

Help mik lẹ với ;-;

\(\left|2x-3y\right|+\left|2y+3z\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-3y=0\\2y+3z=0\\x+y+z=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=3y\\3z=-2y\\x+y+z=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3y}{2}\\z=\dfrac{-2y}{3}\\x+y+z=0\end{matrix}\right.\)

\(\Rightarrow x=y=z=0\)