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3 years ago

How to solve 19,20,21??????

Mathematics

1 answer:

Doss [256]3 years ago

3 0

Answer:

19. undefined, thus no answer

20. yes, answer is 0

21. yes, answer is 5

Step-by-step explanation:

all requires application on the laws and special cases of logarithm

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3 years ago

Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

worty [1.4K]

Answer:

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.

So let's see what happens if we gather the terms together that have the same variable squared together.

$x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18$

Factor the variable squared terms out of each binomial pairing:

$x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18$

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

$6+xyz+xyz+xyz=18$

Combine like terms:

$6+3xyz=18$

Subtract 6 on both sides:

$3xyz=12$

Divide both sides by 3:

$xyz=4$

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3 years ago

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3 years ago

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