A. x - y
B. y - x
C. (xy)^2
D. -(xy)^2
Simplify the expression: (x^2 - y^2) / (x^-2 - y^-2)?
2016-07-06 4:06 am
回答 (12)
2016-07-06 4:14 am
✔ 最佳答案
(x² - y²) / (x⁻² - y⁻²)= (x² - y²) / [(1 / x²) - (1 / y²)]
= (x² - y²) / [(y² / x²y²) - (x² / x²y²)]
= (x² - y²) / [(y² - x²) / x²y²]
= (x² - y²) / [-(x² - y²) / x²y²]
= (x² - y²) × [-x²y² / (x² - y²)]
= -x²y²
= -(xy)²
...... The answer is : D. -(xy)²
2016-07-06 6:40 am
Put n = x^2 - y^2 = (x+y)(x-y). Put d = x^-2 -y^-2 = [(1/x) + (1/y)][(1/x) - (1/y)] = - (x+y)(x-y)/(xy)^2. Then
N/D = - (xy)^2 and choice D. pertains.
N/D = - (xy)^2 and choice D. pertains.
2016-07-06 4:11 am
I would not tell you the answer but I can tell you how to do that.
You may know the identities of Algebraic expressions.
In the first term here it is x^2-x^2 which can be written as x^2-(x+y)100+xy
The second can be done as same. Goodluck it's easy to do.
You may know the identities of Algebraic expressions.
In the first term here it is x^2-x^2 which can be written as x^2-(x+y)100+xy
The second can be done as same. Goodluck it's easy to do.
2016-07-06 7:40 pm
(x^2)((x^2) - (y^2))/(1 - (x^2)(y^(-2)))
(y^2)(x^2)*((x^2) - (y^2))/((y^2) - (x^2))
-((xy)^2)
(y^2)(x^2)*((x^2) - (y^2))/((y^2) - (x^2))
-((xy)^2)
2016-07-06 8:19 am
(x^2 - y^2) / (x^-2 - y^-2)
(x^2 - y^2) / [(y^2 - x^2) / x^2y^2]
x^2y^2(x^2 - y^2) / -(x^2 - y^2)
= -x^2y^2
= -(xy)^2
Answer choice:
D. -(xy)^2
(x^2 - y^2) / [(y^2 - x^2) / x^2y^2]
x^2y^2(x^2 - y^2) / -(x^2 - y^2)
= -x^2y^2
= -(xy)^2
Answer choice:
D. -(xy)^2
2016-07-06 7:47 am
(x^2 - y^2) / (x^-2 - y^-2)
= (x^2 - y^2) / (1/x^2 - 1/y^2) ---> use the power rules, a^-n = 1/a^n. The power becomes positive when it's in the denominator.
now cross multiply to get a common denominator x^2y^2
= (x^2 - y^2) / [(y^2 - x^2)/x^2y^2]
= (x^2 - y^2) / [-(x^2 - y^2) /x^2y^2] ---> factored out a negative so you can cancel the factors.
= [(x^2y^2)*(x^2 - y^2)]/-(x^2 - y^2) --> cancel common factors.
= -(x^2y^2)
= -(xy)^2
answer is D
= (x^2 - y^2) / (1/x^2 - 1/y^2) ---> use the power rules, a^-n = 1/a^n. The power becomes positive when it's in the denominator.
now cross multiply to get a common denominator x^2y^2
= (x^2 - y^2) / [(y^2 - x^2)/x^2y^2]
= (x^2 - y^2) / [-(x^2 - y^2) /x^2y^2] ---> factored out a negative so you can cancel the factors.
= [(x^2y^2)*(x^2 - y^2)]/-(x^2 - y^2) --> cancel common factors.
= -(x^2y^2)
= -(xy)^2
answer is D
參考: grade 10 student.
2016-07-06 5:56 am
x^-2 - y^-2 = 1/x^2 - 1/y^2 = (y^2- x^2)/(xy)^2 = -(x^2-y^2)/(xy)^2
(x^2- y^2)/[-(x^2-y^2)/(xy)^2)] = 1/(-1/(xy)^2) = -(xy)^2
Answer: D.
(x^2- y^2)/[-(x^2-y^2)/(xy)^2)] = 1/(-1/(xy)^2) = -(xy)^2
Answer: D.
2016-07-06 5:18 am
(x^2 - y^2)/(x^-2 - y^-2)
= (x^2 - y^2)/((1/x^2) - (1/y^2))
= (x^2 - y^2)/((y^2 - x^2)/(x^2y^2))
= (x^2 - y^2) * ((x^2y^2)/(y^2 - x^2))
= (x^2 - y^2) * ((x^2y^2)/-1(x^2 - y^2))
= (x^2y^2)/-1
= -x^2y^2
= -(xy)^2. The answer is D).
= (x^2 - y^2)/((1/x^2) - (1/y^2))
= (x^2 - y^2)/((y^2 - x^2)/(x^2y^2))
= (x^2 - y^2) * ((x^2y^2)/(y^2 - x^2))
= (x^2 - y^2) * ((x^2y^2)/-1(x^2 - y^2))
= (x^2y^2)/-1
= -x^2y^2
= -(xy)^2. The answer is D).
2016-07-06 4:17 am
(x^2 - y^2) / (x^-2 - y^-2)
= (x² - y²) / (1/x² - 1/y²)
= (x² - y²) / [(y² - x²)/x²y²]
= (x² - y²) / [-(x² - y²) /x²y²]
= - x²y²
= -(xy)²
= (x² - y²) / (1/x² - 1/y²)
= (x² - y²) / [(y² - x²)/x²y²]
= (x² - y²) / [-(x² - y²) /x²y²]
= - x²y²
= -(xy)²
2016-07-08 10:11 pm
D
2016-07-07 4:44 pm
D
2016-07-07 2:30 pm
yes, i agree with above answer.
Ans::: -x^2 y^2
Ans::: -x^2 y^2
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