Question

(x+y)^-5(x+y)^2/(x+y)^-2(x+y)^-1

Answer 1

Ответ:

1

Пошаговое объяснение:

(x+y)^-5(x+y)^2/(x+y)^-2(x+y)^-1

(x+y)^-3/(x+y)^-3

1

Answer 2

Если условие такое:

$\frac{(x+y)^{-5}(x+y)^2}{(x+y)^{-2}(x+y)^{-1}}=$ $(x+y)^{-5+2-(-2)-(-1)}=(x+y)^{-5+2+2+1}=(x+y)^{0}=1$

то

Ответ: 1

А если условие такое:

$(x+y)^{-5}*\frac{(x+y)^2}{(x+y)^{-2}}*(x+y)^{-1}=(x+y)^{-5+2-(-2)-1}=(x+y)^{-2}=\frac{1}{(x+y)^2}$

то

Ответ:   $\frac{1}{(x+y)^{2}}$

Answer 3

Let's simplify the expression $\frac{(x+y)^{-5}(x+y)^2}{(x+y)^{-2}(x+y)^{-1}}$.

First, we can use the properties of exponents to combine the terms with the same base:

$\frac{(x+y)^{-5}(x+y)^2}{(x+y)^{-2}(x+y)^{-1}} = (x+y)^{(-5 + 2)} \cdot (x+y)^{(-2 - 1)}.$

Now, simplify the exponents:

$= (x+y)^{-3} \cdot (x+y)^{-3}.$

Next, use the rule $a^{-m} = \frac{1}{a^m}$:

$= \frac{1}{(x+y)^3} \cdot \frac{1}{(x+y)^3}.$

Now, apply the rule $a^m \cdot a^n = a^{m+n}$:

$= \frac{1}{(x+y)^{3+3}}.$

Combine the exponents:

$= \frac{1}{(x+y)^6}.$

So, the simplified form of the given expression is $\frac{1}{(x+y)^6}$.