TỪ GT ⇒(X2 -YZ)Y(1-XZ) = X(1- YZ)(Y2 - XZ) ⇔X2Y- X3YZ-Y2Z+XY2Z2 = XY2...
Bài 5:
Từ GT
⇒
(x
2
-yz)y(1-xz) = x(1- yz)(y
2
- xz)
⇔
x
2
y- x
3
yz-y
2
z+xy
2
z
2
= xy
2
-x
2
z - xy
3
z +x
2
yz
2
⇔
x
2
y- x
3
yz - y
2
z+ xy
2
z
2
- xy
2
+x
2
z + xy
3
z - x
2
yz
2
= 0
⇔
xy(x-y) +xyz(yz +y
2
- xz - x
2
)+z(x
2
- y
2
) = 0
⇔
xy(x-y) - xyz(x -y)(x + y +z)+z(x - y)(x+y) = 0
⇔
(x -y)
[xy
−
xyz
(
x
+
y
+
z
)
+
xz
+
yz
]= 0
Do x - y
≠
0 nên xy + xz + yz - xyz ( x + y + z) = 0
Hay xy + xz + yz = xyz ( x + y + z) (đpcm)