Question

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Answer 1

Răspuns:

$x - \frac{2^{500}+2^{501} }{2^{502} } = \frac{4^{100}+4^{101} }{4^{102} }\\\\<=> x= \frac{4^{100}+4^{101} }{4^{102} } + \frac{2^{500}+2^{501} }{2^{502} } \\\\<=> x= \frac{4^{100} }{4^{102} } + \frac{4^{101} }{4^{102} } + \frac{2^{500} }{2^{502} } + \frac{2^{501} }{2^{502} } \\\\<=> x= 4^{-2} + 4^{-1} + 2^{-2} + 2^{-1}\\\\<=> x= \frac{1}{16} + \frac{1}{4} + \frac{1}{4} + \frac{1}{2}\\\\<=> x= \frac{1}{16} + \frac{1}{2} + \frac{1}{2} \\\\<=> x= \frac{1}{16} + 1\\\\<=> x= \frac{17}{16}$

Answer 2

$\bf x - \dfrac{2^{500}+ 2^{501}}{2^{502} }= \dfrac{4^{100}+ 4^{101}}{4^{102} }$

$\bf x = \dfrac{4^{100}+ 4^{101}}{4^{102}} + \dfrac{2^{500}+ 2^{501}}{2^{502}}$

$\bf x = \dfrac{(2^{2})^{100}+ (2^{2})^{101}}{(2^{2})^{102}} + \dfrac{2^{500}\cdot(1+ 2^{1})}{2^{502}}$

$\bf x = \dfrac{2^{2\cdot 100}+ 2^{2 \cdot 101}}{2^{2\cdot 102}} + \dfrac{2^{500}\cdot(1+ 2)}{2^{502}}$

$\bf x = \dfrac{2^{200}+ 2^{202}}{2^{204}} + \dfrac{2^{500}\cdot 3}{2^{502}}$

$\bf x = \dfrac{2^{200}\cdot(1+ 2^{2})}{2^{204}} + \dfrac{2^{500}\cdot 3}{2^{502}}$

$\bf x = \dfrac{2^{200}\cdot(1+ 4)}{2^{204}} + \dfrac{\not2^{500}\cdot 3}{\not2^{502}}$

$\bf x = \dfrac{2^{200}\cdot5}{2^{204}} + \dfrac{ 3}{2^{2}}$

$\bf x = \dfrac{\not2^{200}\cdot5}{\not2^{204}} + \dfrac{ 3}{4}}$

$\bf x = \dfrac{5}{2^{4}} + \dfrac{ 3}{4}}$

$\bf x = \dfrac{5}{16}} + \dfrac{ 3\cdot 4}{4\cdot4}}$

$\bf x = \dfrac{5}{16}} + \dfrac{ 12}{16}}$

$\boxed{\bf x = \dfrac{17}{16}}$

Cateva formule pentru puteri

a⁰ = 1 sau 1 = a⁰

(aⁿ)ᵇ = aⁿ ˣ ᵇ sau aⁿ ˣ ᵇ = (aⁿ) ᵇ

aⁿ · aᵇ = (a · a) ⁿ ⁺ ᵇ  sau  (a · a) ⁿ ⁺ ᵇ = aⁿ · aᵇ

aⁿ : aᵇ = (a : a) ⁿ ⁻ ᵇ sau (a : a) ⁿ ⁻ ᵇ = aⁿ : aᵇ

aⁿ · bⁿ = (a · b)ⁿ sau (a · b)ⁿ = aⁿ · bⁿ

aⁿ : bⁿ = (a : b)ⁿ sau (a : b)ⁿ = aⁿ : bⁿ

(- a)ⁿ,unde n este o putere impara (-a)ⁿ=(-a)ⁿ

(- a)ⁿ,unde n este o putere para (-a)ⁿ = aⁿ