Lời giải:
Ta có: \(\left\{\begin{matrix} xy+x+y=3\\ yz+y+z=8\\ zx+z+x=15\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+1)(y+1)=4\\ (y+1)(z+1)=9\\ (z+1)(x+1)=16\end{matrix}\right.(1)\)
Nhân 3 vế với nhau:
\(\Rightarrow [(x+1)(y+1)(z+1)]^2=4.9.16\)
\(\Leftrightarrow (x+1)(y+1)(z+1)=\pm 24\)
Nếu \((x+1)(y+1)(z+1)=24(2)\)
Từ \((1),(2)\Rightarrow \left\{\begin{matrix} z+1=6\\ x+1=\frac{8}{3}\\ y+1=\frac{3}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{5}{3}\\ y=\frac{1}{2}\\ z=5\end{matrix}\right.\)
Do đó, \(P=x+y+z=\frac{43}{6}\)
Nếu
\((x+1)(y+1)(z+1)=-24(3)\)
Từ $(1);(3)$ suy ra \(\left\{\begin{matrix} z+1=-6\\ x+1=\frac{-8}{3}\\ y+1=\frac{-3}{2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} z=-7\\ x=-\frac{11}{3}\\ y=\frac{-5}{2}\end{matrix}\right.\)
Do đó, \(P=x+y+z=-\frac{79}{6}\)
Đề sai.Sửa đề: \(xy+x+y=3\)
\(\left\{{}\begin{matrix}xy+x+y=3\\yz+y+z=8\\xz+x+z=15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}xy+x+y+1=4\\yz+y+z+1=9\\xz+x+z+1=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y\left(x+1\right)+1\left(x+1\right)=4\\y\left(z+1\right)+1\left(z+1\right)=9\\x\left(z+1\right)+1\left(z+1\right)=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\left(y+1\right)\left(z+1\right)=9\\\left(x+1\right)\left(z+1\right)=16\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(y+1\right)\left(z+1\right)\left(x+1\right)\left(z+1\right)=576\)
\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=576\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=24\)
Đến đây chịu :v
Ta có:
xy+y+z=3
yz+y+z=8
xz+x+z=15
=> xy+y+z+yz+y+z+xz+x+z = 3+8+15= 26
=>xy+yz+xz + 2(y+z+x) = 26
Vì x,y,z >0 => xy>0;yz>0;xz>0 và xy+yz+xz > y+x+z
=> xy+yz+xz =3 thì y+x+z =11,5 (không hợp lý)
=> xy+yz+xz >3 thì y+z+x <11,5
Mà xy +yz +xz > y+x+z. Do đó y+z+x <3 hoặc =3
=> y+z+x =1;2;3
Không có trường hợp x+z+y =0
Mình thật sự không biết cách giải :>> Giải bừa thôi ạ
