\(-\frac{3}{5}xyz^2\cdot\frac{1}{3}xy\cdot\left(-\frac{1}{4}\right)x^5yz\)

\(=\left(-\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{-1}{4}\right)\left(x\cdot x\cdot x^5\right)\left(y\cdot y\cdot y\right)\left(z^2\cdot z\right)\)

\(=\frac{1}{20}x^7y^3z^3\)

\(=xyz^2\)

đúng không

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

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

=\(xyz^2\)

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

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

=\(xyz^2.1\)

= \(xyz^2\)

(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0

(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0

(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0

(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0

Suy ra x+y+z =0 

x+y = -z

y+z = -x

x+z = -y

B = -16 + (-3) +2038 = 2019

Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)

+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)

\(=-16-3+2038=2019\)

+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)

\(=32+6-4076=-4038\)

Cho em hỏi chút ạ, trường hợp  x=y=z suy ra ntn ạ?

Bài b nhé bạn!

\(\hept{\begin{cases}\frac{xyz}{x+y}=2\\\frac{xyz}{y+z}=\frac{6}{5}\\\frac{xyz}{x+z}=\frac{3}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{1}{yz}+\frac{1}{xz}=\frac{1}{2}\\\frac{1}{xz}+\frac{1}{xy}=\frac{5}{6}\\\frac{1}{xy}+\frac{1}{yz}=\frac{2}{3}\end{cases}}\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=\frac{\frac{1}{2}+\frac{5}{6}+\frac{2}{3}}{2}=1\)

Trừ lại từng phương trình trong hệ:

\(\hept{\begin{cases}\frac{1}{xy}=\frac{1}{2}\\\frac{1}{yz}=\frac{1}{6}\\\frac{1}{xz}=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}xy=2\\yz=6\\xz=3\end{cases}\Rightarrow xyz=\sqrt{2.6.3}=6}\)

Chia lại từng phương trình trong hệ mới, được:

\(\hept{\begin{cases}z=3\\x=1\\y=2\end{cases}}\)

Vậy \(\left(x;y;z\right)=\left(1;2;3\right)\)

Xong rồi đó!!!

Ta có: x³ + y³ + z³ = 3xyz

x³ + y³ + z³ - 3xyz = 0

x³ + 3x²y + 3xy² + y³ + z³ - 3xy(x + y) - 3xyz = 0

(x + y)³ + z² - 3xy(x + y + z) = 0

(x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy) = 0

(x + y + z)(x² + y² + z² - xz - yz - xy) = 0

=> x + y + z = 0 hoặc x² + y² + z² - xz - yz - xy = 0

+) Với x + y + z = 0 

<=> x + y = -z, x + z = -y, y + z = -x

Thay x + y = -z, x + z = -y, y + z = -x vào P, ta có:

\(P=\frac{xyz}{\left(-z\right)\left(-x\right)\left(-y\right)}=-1\)

+) Với x² + y² + z² - xz - yz - xy = 0

=> 2x² + 2y² + 2z² - 2xz - 2yz - 2xy = 0

=> (x² - 2xy + y²) + (x² - 2xz + z²) + (y² - 2yz + z²) = 0

=> (x - y)² + (x - z)² + (y - z)² = 0

=> (x - y)² = 0 và (x - z)² = 0 và (y - z)² = 0

=> x = y và x = z và y = z

=> x = y = z

Thay x = y = z vào P, ta có:

\(P=\frac{xxx}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{x^3}{\left(2x\right)^3}=\frac{x^3}{8x^3}=\frac{1}{8}\)

Đặt \(\left ( \frac{1}{xy},\frac{1}{yz},\frac{1}{xz} \right )=(a,b,c)\)

\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} b+c=\frac{1}{2}\\ c+a=\frac{5}{6}\\ a+b=\frac{2}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2b=\frac{2}{3}+\frac{1}{2}-\frac{5}{6}\\ 2c=\frac{1}{2}+\frac{5}{6}-\frac{2}{3}\\ 2a=\frac{5}{6}+\frac{2}{3}-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} b=\frac{1}{6}\\ c=\frac{1}{3}\\ a=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} yz=6\\ xz=3\\ xy=2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=3\end{matrix}\right.\)

\(\left\{\begin{matrix}\frac{x+y}{xyz}=\frac{1}{2}\\\frac{y+z}{xyz}=\frac{5}{6}\\\frac{x+z}{xyz}=\frac{2}{3}\end{matrix}\right.\).Cộng theo vế ta có:

\(\frac{x+y+y+z+x+z}{xyz}=\frac{1}{2}+\frac{5}{6}+\frac{2}{3}=2\)

\(\Leftrightarrow\frac{2\left(x+y+z\right)}{xyz}=2\Rightarrow2\left(x+y+z\right)=2xyz\)

\(\Leftrightarrow x+y+z=xyz\). Thay vào hệ đầu ta có:

\(\left\{\begin{matrix}\frac{x+y}{x+y+z}=\frac{1}{2}\\\frac{y+z}{x+y+z}=\frac{5}{6}\\\frac{x+z}{x+y+z}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2\left(x+y\right)=x+y+z\\6\left(y+z\right)=5\left(x+y+z\right)\\3\left(x+z\right)=2\left(x+y+z\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}2\left(x+y\right)=x+y+z\\\frac{6}{5}\left(y+z\right)=x+y+z\\\frac{3}{2}\left(x+z\right)=x+y+z\end{matrix}\right.\)

\(\Leftrightarrow2x+2y=\frac{6}{5}y+\frac{6}{5}z=\frac{3}{2}x+\frac{3}{2}z=x+y+z\)\(\Leftrightarrow\left\{\begin{matrix}y=2x\\z=3x\end{matrix}\right.\)

giải tùm lum quá ra cái j ko bt