Question

Cho \(x,y,z>=0\)thõa \(x+y+z=1\)và \(x^3+y^3+z^3=\frac{1}{9}\)tính \(C=\left(1-x\right)\left(1-y\right)\left(1-z\right)\)làm ơn giúp mình với, cảm ơn các bạn nhìu!

Answer 1

Thay x+y+z=1 vào biểu thức C, ta được:

\(C=\left(x+y+z-x\right)\left(x+y+z-y\right)\left(x+y+z-z\right)\)

\(C=\left(y+z\right)\left(z+x\right)\left(x+y\right)=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta có: \(x^3+y^3+z^3=\frac{1}{9}\Leftrightarrow\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)

Thay x+y+z=1. Suy ra \(1-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)

\(\Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9}\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9.3}=\frac{8}{27}\)

\(\Rightarrow C=\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{27}.\)

ĐS:...