Question

x(y+z)=32 , y(x+z)=27 , z(y+x)=35 hỏi xyz=?

Answer 1

Ta có:

\(x\left(y+z\right)=32\Rightarrow xy+xz=32\)

\(y\left(x+z\right)=27\Rightarrow xy+yz=27\)

\(z\left(y+x\right)=yz+xz=35\)

\(\Rightarrow xy+xz+xy+yz+yz+xz=32+27+35\)

\(\Rightarrow2\left(xy+yz+zx\right)=94\)

\(\Rightarrow xy+yz+zx=47\)

Mà \(xy+yz=27\)

\(\Rightarrow27+zx=47\)

\(\Rightarrow zx=20\)

Tương tự ta được : \(xy=12\) ; \(yz=15\)

\(\Rightarrow zx.xy=20.12\)

\(\Rightarrow x^2.yz=240\)

Mà \(yz=15\)

\(\Rightarrow x^2=240:15\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\pm4\)

+)Nếu \(x=4\Rightarrow xyz=4.yz=4.15=60\)

+)Nếu \(x=-4\Rightarrow xyz=-4.yz=-4.15=-60\)

Vậy \(xyz=60;xyz=-60\)

Answer 2

xyz = 60