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Addition and Subtraction of Algebraic Expression

If x+y+z=0,...

Question

If $x + y + z = 0$ , then value of $\frac{(x + y) (y + z) (z + x)}{x y z} + 11$ is ?

A

x + 11

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B

y + 11

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C

z + 11

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D

10

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Solution

The correct option is C $10$
$x + y + z = 0 ⟺ x + y = - z ⟺ y + z = - x ⟺ z + x = - y$
Substituting the above in the expression given:
$\frac{(- z) (- x) (- y)}{x y z} + 11 = - 1 + 11 = 10$

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Similar questions

x > y > z > 0

, then prove that

{cot}^{- 1} (\frac{1 + x y}{x - y}) + {cot}^{- 1} (\frac{1 + x y}{y - z}) + {cot}^{- 1} (\frac{1 - z x}{z - x}) = π

x + y + z = 0

,then prove that

\frac{x y z}{(x + y) (y + z) (z + x)} = - 1

x \neq - y, y \neq - z, z \neq - x

(x + y + z = 0)

(x + y) (y + z) (z + x)

(x, y, z)

be an arbitrary point lying on a plane

P

which passes through the points

(42, 0, 0),

(0, 42, 0)

(0, 0, 42),

then the value of the expression

3 + \frac{x - 11}{(y - 19)^{2} (z - 12)^{2}} + \frac{y - 19}{(x - 11)^{2} (z - 12)^{2}} + \frac{z - 12}{(x - 11)^{2} (y - 19)^{2}} - \frac{x + y + z}{14 (x - 11) (y - 19) (z - 12)}

If x+y+z=6, xy+xz+yz=11, xyz=6 thenequals

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