P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)

1/2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)

1/2P=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

1/2P=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

1/2P=(5^8-1)(5^8+1)(5^16+1)

1/2P=(5^16-1)(5^16+1)

1/2P=5^32-1

P=(5^32-1):1/2=2(5^32-1)

a)x¹.x².x³....x²⁰¹⁷

=x^{(1+2+3+...+2017)}

ta có :

1+2+3+...+2017= [(2017-1):1+1]:2.2018

= 2035153

=>x¹.x².x³....x²⁰¹⁷=x^2035153

b) x¹.x⁴.x⁷....x¹⁹⁶³

=x^{(1+4+7+...+1963)}

ta có:

1+4+7+....+1963=[(1963-1):3+1]:2.1964

=643210

=>x¹.x⁴.x⁷....x¹⁹⁶³=x⁶⁴³²¹⁰

= 0 từ trên xuống dưới

(x+9) ⋮ (x-1)

Ta có:

(x+9) ⋮ (x-1)

<=>[(x-1)+10] ⋮ (x-1)

Vì x-1 ϵ Ư(10) = {-10; -5; -2; -1; 0; 1; 2; 5; 10}

=>Ta có bảng:

Vì x là số tự nhiên

=> x ϵ {0; 1; 2; 3; 6; 11}

Hok tốt

4⁶ . 9⁵ + 6⁹ . 120 / -8⁴ . 3¹² - 6¹¹

= (2²)⁶ . (3²)⁵ + (2.3)⁹ . 2³ . 3 . 5 / -[ (2³)⁴ . 3¹² + (2.3)¹¹]

= 2¹² . 3¹⁰ + 2⁹ . 3⁹ . 2³ . 3 . 5 / -( 2¹² . 3¹² + 2¹¹ . 3¹¹)

= 2¹² . 3¹⁰ + 2¹² . 3¹⁰ . 5 / -[ 2¹¹ . 3¹¹ . ( 2.3 + 1)]

= 2¹² . 3¹⁰ . ( 1 + 5) / -[ 2¹¹ . 3¹¹ . ( 6 + 1)]

= 2¹² . 3¹⁰ . 6 / -( 2¹¹ . 3¹¹ . 7)

= 2¹³ . 3¹¹ / -( 2¹¹ . 3¹¹ . 7)

= 2² / -7 = 4/-7

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-2\times\sqrt{6}\times3+9}\)\(+\sqrt{\left(2\sqrt{6}\right)^2-2\times2\sqrt{6}\times3+9}\)

\(=\sqrt{\left(\sqrt{6}-3\right)^2}\)\(+\sqrt{\left(2\sqrt{6}-3\right)^2=\sqrt{6}-3+2\sqrt{6}-3=3\sqrt{6}-3}\)

\(V\text{ậya=-3;b=3a=-3;b=3}\)

Số các số có 4 chữ số là:

              (9999-1000):1+1=9000(số)