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ОтветитьMethod 3:
Expand the squares and divide by the denominators gives
x/y + 2 + y/x +x/z + 2 + z/x + y/z + 2 + z/y
= 6 + (y + z) / x + (x + z) / y + (x + y) / z
= 6 + (-x) / x + (-y) / y + (-z) / z
= 6 - 3
= 3
x+y+z=0
Let a=[(x+y)²/xy]+[(x+z)²/xz]
+[(y+z)²/yz]
=(z²/xy)+(y²/xz)+(x²/yz)
=(x³+y³+z³)/xyz
0=(x+y+z)³ as x+y+z=0
=x³+y³+z³+3(x+y+z)(xy+yz+zx)
-3xyz
=x³+y³+z³-3xyz
(x³+y³+z³)/xyz=3 --> a=3