Question

How to solve this? Is the question correct? ty :)

Answer 1

Answer:

The proposition is true.

Step-by-step explanation:

Now we proceed to demostrate that expression given is true by algebraic means:

1) $\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1} }{y^{-1}}$ Given

2) $\frac{y^{-1}\cdot (x^{-1}+y^{-1})+x^{-1}\cdot (x^{-1}-y^{-1})}{x^{-1}\cdot y^{-1}}$   $\frac{a}{b} + \frac{c}{d} = \frac{a\cdot d+b\cdot c}{b\cdot d}$

3) $\frac{x^{-1}\cdot y^{-1}+y^{-2}+x^{-2}-x^{-1}\cdot y^{-1}}{x^{-1}\cdot y^{-1}}$ Distributive property/$a^{b}\cdot a^{c} = a^{b+c}$

4) $\frac{x^{-2}+y^{-2}}{(x\cdot y)^{-1}}$ Commutative, associative and modulative properties/Existence of additive inverse/$a^{b}\cdot c^{b} = (a\cdot c)^{b}$

5) $[(x\cdot y)^{-1}]^{-1}\cdot (x^{-2}+y^{-2})$ Commutative property/Definition of division

6) $(x\cdot y)\cdot (x^{-2}+y^{-2})$         $(x^{-1})^{-1}$

7) $x^{-1}\cdot y + x\cdot y^{-1}$ Distributive property/Associative property/$a^{b}\cdot a^{c} = a^{b+c}$

8) $(x^{-1}\cdot y^{-1})\cdot y^{2} + (x^{-1}\cdot y^{-1})\cdot x^{2}$  Modulative property/Existence of additive inverse/$a^{b}\cdot a^{c} = a^{b+c}$

9) $(x^{-1}\cdot y^{-1})\cdot (x^{2}+y^{2})$ Distributive property

10) $(x\cdot y)^{-1}\cdot (x^{2}+y^{2})$ $a^{b}\cdot c^{b} = (a\cdot c)^{b}$

11) $\frac{x^{2}+y^{2}}{x\cdot y}$ Commutative property/Definition of division/Result