=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=...=2^32-1

nhân hết ra là xong:))

bài về nhà hs phải tự làm

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

 =2.(5^4-1).(5^4+1).(5^8+1).(5^16+1)

= 2.(5^8-1).(5^8+1).(5^16+1)

= 2.(5^16-1).(5^16+1)

= 2.(5^32-1)

 1)P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1) 
=(5^4-1)(5^4+1)(5^8+1)(5^16+1) 
=(5^8-1)(5^8+1)(5^16+1) 
=(5^16-1)(5^16+1) 
=5^32-1 
=> P = (5^32-1)/2 

a, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 thì 
P = [x/(x^2 - 25)  -  (x - 5)/(x^2 + 5x)]  : (2x - 5)/(x^2 + 5x) + x/(x - 5)
<=>P = [x/(x - 5)(x + 5)  -  (x - 5)/x(x+5)] . x(x + 5)/(2x - 5) + x/(x - 5)
=> P = [x^2 - (x - 5)^2]/x(x - 5)(x + 5) . x(x + 5)/(2x - 5) + x/(x - 5)
<=> P = (x - x + 5)(x + x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5(2x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5/(x - 5) + x/(x - 5)
<=> P = (5 + x)/(x - 5)
b, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 (x ∈ Z) thì P ∈ Z <=> (5 + x)/(x - 5) ∈ Z
<=> (x - 5 + 10)/(x - 5) ∈ Z
<=> 1 + 10/(x - 5) ∈ Z
<=> 10/(x - 5) ∈ Z
<=> (x - 5) ∈ Ư(10)
<=> x - 5 = 10  <=> x = 15 (TM)
hoặc x - 5 = -10 <=> x = -5  (TM)
hoặc x - 5 = 5  <=> x = 10  (TM)
hoặc x - 5 = -5 <=> x = 0  (TM)
hoặc x - 5 = 2  <=> x = 7  (TM)
hoặc x - 5 = -2  <=> x = 3  (TM)
hoặc x - 5 = -1  <=> x = 4  (TM)
hoặc x - 5 = 1  <=> x = 6  (TM)
Vậy x ∈ {-5,0,3,4,6,7,10,15} thì P ∈ Z

12x²  + 8x - 9x - 6 - 12x² + 30x + 2x -5 + 1= 31x - 10

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

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

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

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

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

2P = 5^32 - 1

P = (5^32 - 1) : 2

\(P=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Rightarrow2P=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2P=5^{32}-1\)

\(\Leftrightarrow P=\frac{5^{32}-1}{2}\)

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy