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mote1985 [20]
3 years ago
14

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

Mathematics
1 answer:
bezimeni [28]3 years ago
4 0

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

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Induction step: show that the assumption implies (<em>k</em> + 1)² - (<em>k</em> + 1) is also even. We have

(<em>k</em> + 1)² - (<em>k</em> + 1) = <em>k</em> ² + 2<em>k</em> + 1 - <em>k</em> - 1

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… = 2<em>m</em> + 2<em>k</em>

… = 2 (<em>m</em> + <em>k</em>)

which is clearly even. QED

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Step-by-step explanation:

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