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Zorluk: ZorEvaluating Algebraic Expressions

If x=−3x = -3, y=−13y = -\frac{1}{3}, and z=16z = 16, what is the value of the algebraic expression x3y−2+3z1/2(xy+2)3−32z1/2\frac{x^3 y^{-2} + 3 z^{1/2}}{(xy + 2)^3 - \frac{3}{2} z^{1/2}}?

Cevap: -11

Cevap

The evaluated value of the expression is -11.
Substituting the given values into the numerator yields (−3)3(−13)−2+3(16)1/2=(−27)(9)+3(4)=−243+12=−231(-3)^3 \left(-\frac{1}{3}\right)^{-2} + 3(16)^{1/2} = (-27)(9) + 3(4) = -243 + 12 = -231. Substituting into the denominator yields ((−3)(−13)+2)3−32(16)1/2=(1+2)3−32(4)=27−6=21\left((-3)\left(-\frac{1}{3}\right) + 2\right)^3 - \frac{3}{2}(16)^{1/2} = (1 + 2)^3 - \frac{3}{2}(4) = 27 - 6 = 21. Dividing −231-231 by 2121 produces −11-11.

Adım Adım Çözüm

1
Evaluate the terms in the numerator
Numerator = -231
Calculate x3=(−3)3=−27x^3 = (-3)^3 = -27, y−2=(−13)−2=9y^{-2} = \left(-\frac{1}{3}\right)^{-2} = 9, giving x3y−2=−243x^3 y^{-2} = -243. Then calculate 3z1/2=3(4)=123 z^{1/2} = 3(4) = 12. Adding these gives −243+12=−231-243 + 12 = -231.
2
Evaluate the terms in the denominator
Denominator = 21
Calculate xy=(−3)(−13)=1xy = (-3)\left(-\frac{1}{3}\right) = 1, so (xy+2)3=(1+2)3=27(xy + 2)^3 = (1 + 2)^3 = 27. Then calculate 32z1/2=32(4)=6\frac{3}{2} z^{1/2} = \frac{3}{2}(4) = 6. Subtracting gives 27−6=2127 - 6 = 21.
3
Divide the evaluated numerator by the evaluated denominator
-11
−23121=−11\frac{-231}{21} = -11.

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