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x, y and z are the positive integers such that x > y > z. Which of the following is closest to the product xyz?

1. xy(z-1)
2. (x-1)yz
3. (x-y)xy
4. x(y+1)z

Please describe the solution in detail.

a. $xy(z-1) = xyz - xy$

b. $(x-1)yz = xyz - yz$

c. Not an option as $z$ is missing

d. $x(y+1)z = xyz + xz$

Since $x,y,z >0$, and $x > y > z, yz < xz < xy$

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