Question

Se dau numerele:
x=
$\frac{a + {8}^{1004} }{a + ( {3}^{4} \times {3}^{5} \times {3}^{6} \times ......... \times {3}^{63}) \div 9 }$
, a este nr nat, și y =
$\frac{ {10}^{150} }{2 \times ( 3 + {3}^{2} +........ + {3}^{299} }$
a)Arătați că x<1.
b)Comparați numerele x și y.

Answer 1

Totul sta in niste formule cu sume.

$3^{4} *3^{5} *3^{6} *...*3^{63} =3^{4+5+6+...+63}$

$4+5+6+...+63 = 1+2+3+4+5+6+...+63-1-2-3=63*64:2-1-2-3=2016-6=2010$

$x=\frac{a+8^{1004}} {a+3^{2010}:3^{2}}=\frac{a+8^{1004}} {a+3^{2008}}= \frac{a+8^{1004}} {a+(3^{2})^{1004}}=\frac{a+8^{1004}} {a+9^{1004}}$

$\frac{a+8^{1004}} {a+9^{1004}}<1$ ⇒ $a+8^{1004}<a+9^{1004}$ ⇒ $8^{1004}<9^{1004}$ ⇒ $8<9$ (Adevarat)

$3+3^{2} +...+3^{299} = \frac{3^{300} - 3}{2}$

$y=\frac{10^{150} }{2*\frac{3^{300} - 3}{2}}=\frac{10^{150}}{3^{300} - 3} =\frac{10^{150}}{(3^{2})^{150} - 3}=\frac{10^{150}}{9^{150} - 3}$ ⇒ $y>1$

$y>x$