\(A=x^4+y^4-2x^3-2x^2y^2+x^2-2y^3+y^2\)

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

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

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

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

\(A=x^2-2xy+y^2-2x^2+2xy-2y^2+x^2+y^2\)

\(A=0\)

Xét \(x^3-x^2+x-5=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}\right)=\frac{128}{27}\)

Xét \(y^3-2y^2+2y+4=0\)

\(\Leftrightarrow\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(y-\frac{2}{3}\right)=-\frac{128}{27}\)

Cộng theo vế 2 dòng có dấu <=> ta có:

\(\left(x-\frac{1}{3}\right)^3+\left(y-\frac{2}{3}\right)^3+\frac{2}{3}\left(x-\frac{1}{3}+y-\frac{2}{3}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}+y-\frac{2}{3}\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2\right)+\frac{2}{3}\left(x+y-1\right)=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}\right)=0\)

Dễ thấy: \(\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\left(y-\frac{2}{3}\right)+\left(y-\frac{2}{3}\right)^2+\frac{2}{3}>0\)

\(\Rightarrow x+y-1=0\Rightarrow x+y=1\)

Done !!!

sai hdt roi ban oi

 (x - 1)/2 = (y - 2)/3 = (z - 3)/4  

=> (x - 1)/2 = 2(y - 2)/6 = 3(z - 3)/12 = [(x - 1) - 2(y - 2) + 3(z - 3)]/(2 - 6 + 12) = [(x - 2y + 3z) - 6]/8  

Vì x - 2y + 3z = 14  

=> (x - 1)/2 = (y - 2)/3 = (z - 3)/4 = (14 - 6)/8 = 1  

=> x = 3, y = 5, z = 7 

Vay khi : x+y+z=3+5+7=15