Bài 8 :
a) $\ 2^{x} . 4 = 128$
$\ ⇒ 2^{x} . 2^{2} = 2^{7}$
$\ ⇒ 2^{x + 2} = 2^{7}$
$\ ⇒ x + 2 = 7$
$\ ⇒ x = 5 ∈ N$ (TM)
Vậy $\ x = 5$
b) $\ x^{15} = x$
$\ ⇒ x^{15} - x = 0$
$\ ⇒ x . x^{14} - x = 0$
$\ ⇒ x(x^{14} - 1) = 0$
$\ ⇒ \left[ \begin{array}{l}x=0\\x^{14}-1=0\end{array} \right.$
$\ ⇒ \left[ \begin{array}{l}x=0\\x^{14}=1\end{array} \right.$
$\ ⇒ \left[ \begin{array}{l}x=0 ∈ N (TM)\\x=1 ∈ N (TM)\end{array} \right.$
Vậy $\text{x ∈ { 0 ; 1 }}$
c) $\ 2^{x} . (2^{2})^{2} = (2^{3})^{2}$
$\ ⇒ 2^{x} . 2^{4} = 2^{6}$
$\ ⇒ 2^{x + 4} = 2^{6}$
$\ ⇒ x + 4 = 6$
$\ ⇒ x = 2 ∈ N (TM)$
Vậy $\ x = 2$
d) $\ (x^{5})^{10} = x$
$\ ⇒ x^{50} = x$
$\text{⇒ x ∈ { 0 ; 1 }}$ (theo câu b)
Vậy $\text{x ∈ { 0 ; 1}}$
Bài 9 :
a) Ta có :
$\ a + a^{2} + a^{3} + a^{4} + ... + a^{29} + a^{30}$
$\ = (a + a.a) + (a . a^{2} + a . a^{3}) + ... + (a . a^{28} + a . ^{29})$
$\ = a(a + 1) + a(a^{2} + a^{3}) + ... + a(a^{28} + a^{29})$
$\ = a(a + 1) + a(a^{2} + a^{2} . a) + ... + a(a^{28} + a^{28} . a)$
$\ = a(a + 1) + a . a^{2}(a + 1) + ... + a . a^{28}(a + 1)$
$\ = a(a + 1) + a^{3}(a + 1) + ... + a^{29}(a + 1)$
$\ = (a + 1)(a + a^{3} + ... + a^{29}) \vdots (a + 1) ∨ a ∈ N$ $\ (đpcm)$
