Question

chỉ mình với nhoa :)

Answer 1

Ta có: \(\left(x+y\right)^2=7\)

\(\Rightarrow x^2+2xy+y^2=7\)

\(\Rightarrow\left(x^2+y^2\right)+2xy=7\)

\(\Rightarrow3+2xy=7\)

\(\Rightarrow2xy=4\)

\(\Rightarrow xy=2\) (*)

Thay (*) vào \(xyz=60\) ta có:
\(2z=60\)

\(\Rightarrow z=30\)

Vậy z = 30

Answer 2

30

Answer 3

\(\left(x+y\right)^2=7\Leftrightarrow x^2+2xy+y^2=7\)

Mà \(x^2+y^2=3\Rightarrow2xy=7-3=4\Rightarrow xy=\frac{4}{2}=2\)Ta có: xyz=60=>z=60/2=30

Answer 4

z=30