Aratati că numerele x şi y sunt pătrate perfecte:
x= [2⁹⁰⁰×(2⁶)¹⁰⁰×2 +(64⁴)¹⁰⁰ : 2⁸⁹⁹]² -2³⁰⁰⁷ şi y = 5×(3²⁰⁰² - 3²⁰⁰¹ - 9¹⁰⁰⁰).​

Răspuns:

Explicație pas cu pas:

(64⁴)¹⁰⁰ = 64^400 = (2^6)^400 = 2^2400

x= [2⁹⁰⁰×(2⁶)¹⁰⁰×2 +(64⁴)¹⁰⁰ : 2⁸⁹⁹]² -2³⁰⁰⁷

=(2^900*2^600*2 + 2^2400 : 2^899)^2 - 2^3007

= (2^1501 + 2^1501)^2 - 2^3007

= (2*2^1501)^2 - 2^3007 = (2^1502)^2 - 2^3007 = 2^3004 - 2^3007

= 2^3004*(1 - 2^3) = 2^3004*(1 - 8) = -7*2^3004 nu este patrat perfect

verifica daca ai scris corect exetcitiul; poate nu este 2^3007

_____________

y = 5×(3²⁰⁰² - 3²⁰⁰¹ - 9¹⁰⁰⁰)

= 5*(3^2002 - 3^2001 - 3^2000)

= 5*3^2000*(3^2 - 3 - 1) = 5*3^2000*5 = 5^2*3^2000 = (5*3^1000)^2 patrat perfect

[tex]\begin{aligned} x&= \left[2^{900}\cdot \left(2^6\right)^{100}\cdot 2+\left(64^4\right)^{100}:2^{899}\right]^2-2^{3007} \\ &= \left(2^{900}\cdot 2^{6\cdot 100}\cdot 2+64^{4\cdot 100}:2^{899}\right)^2-2^{3007} \\ &= \left(2^{900}\cdot 2^{600}\cdot 2+64^{400}:2^{899}\right)^2-2^{3007} \\ &= \left(2^{900+600+1}+\left(2^{6}\right)^{400}:2^{899}\right)^2 -2^{3007}\\ &= \left(2^{1501}+2^{6\cdot 400}:2^{899}\right)^2-2^{3007} \\ &= \left(2^{1501}+2^{2400}:2^{899}\right)^2-2^{3007}\\ &= \left(2^{1501}+2^{2400-899}\right)^2-2^{3007} \\ &= \left(2\cdot 2^{1501}\right)^2-2^{3007} \\ &= \left(2^{1501+1}\right)^2-2^{3007} \\ &= \left(2^{1502}\right)^2-2^{3007} \\ &= 2^{1502\cdot 2}-2^{3007} \\ &= 2^{3004}-2^{3007} \\ &= 2^{3004}\cdot \left(1 - 2^{3}\right) \\ &= 2^{3004}\cdot \left(1 - 8\right)\\ &= 2^{3004}\cdot (-7) \\ &=-7\cdot 2^{3004} \to \text{nu este p\u{a}trat perfect} \end{aligned}\\\\[/tex]

[tex]\begin{aligned} y &= 5\cdot \left(3^{2002} - 3^{2001}-9^{1000}\right) \\ &= 5\cdot \left[3^{2002} - 3^{2001}-\left(3^2\right)^{1000}\right] \\ &= 5\cdot \left(3^{2002} - 3^{2001}-2^{2000}\right) \\ &= 5\cdot \left[3^{2000}\cdot \left( 3^2-3^1-1\right)\right]\\ &= 5\cdot 3^{2000}\cdot \left(9-3-1\right)\\ &= 5\cdot 3^{2000}\cdot 5 \\ &= 5^{2}\cdot 3^{2000} \\ &= 5^2\cdot 3^{1000\cdot 2}\\&= 5^{2}\cdot \left(3^{1000}\right)^2 \\ &= \left(5\cdot 3^{1000}\right)^2\to \text{este p\u{a}trat perfect}\end{aligned}[/tex]