value of expression

por **stuart clark** » Seg Mai 30, 2011 00:31

If

are real no. such that $\left\{\begin{array}{c} x+y+z=2\\ x^2+y^2+z^2=16\\ xyz=1\end{array}\right.$ .Then Calculate value of $\displaystyle \frac{1}{xy+2z}+\frac{1}{yz+2x}+\frac{1}{zx+2y}$

por **FilipeCaceres** » Seg Mai 30, 2011 01:34

Note that

,

, and

.

From here, notice that xy + 2z = (x-2)(y-2)

,

, and

.

It should be clear that what we need to do is construct a polynomial with $x-2,\, y-2,\, z-2$ as roots.

Firstly, construct a polynomial with x, y, z as roots. From x + y + z = 2

and $x^{2} + y^{2} + z^{2} = 16$ , get the equation xy + yz + xz = -6

.

Thus a cubic with roots x, y, z is $a^{3} - 2a^{2} - 6a - 1$ .

A cubic with roots $x-2,\, y-2,\, z-2$ is $(a+2)^{3} - 2(a+2)^{2} - 6(a+2) - 1 = a^{3} + 4a^{2} - 2a - 13$ .

$\frac{1}{(x-2)(y-2)} + \frac{1}{(y-2)(z-2)} + \frac{1}{(x-2)(z-2)} = \frac{-4}{13}$

Answer: $\boxed{\frac{-4}{13}}$

por **stuart clark** » Seg Mai 30, 2011 06:27

Thanks FilipeCaceres

value of expression

value of expression

value of expression

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