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\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\cdot\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
\(A=8.\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^8-1\right)....\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
A = 8.(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)
= (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)
= (3¹⁶ - 1)(3¹⁶ + 1)
= 3³² - 1
a) Ta có : 2005.2007 = (2006 - 1)(2006 + 1) = 20062 - 12 = 20062 - 1 ( cái khúc này sửa : 2005.2001 thành 2005.2007)
Mà B = 20062
=> 20062 - 1 < 20062
=> A < B
b) Ta có : B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (24 - 1)(24 + 1)(28 + 1)(216 + 1)
B = (28 - 1)(28 + 1)(216 + 1) = (216 - 1)(216 + 1) = 232 - 1
Mà C = 232
=> B < C
c) Tương tự như câu b
\(\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}-1\right)\\ ...\\ 2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\\ 2A=3^{128}-1\)
Vậy \(A=\dfrac{3^{128}-1}{2}.\)
Bài 1:
a.
$=(x^3+2^3)-(x^3-2)=2^3+2=10$
b.
$=(x^2+10x+25)-4x(4x^2+12x+9)-(2x-1)(x^2-9)$
$=x^2+10x+25-16x^3-48x^2-36x-(2x^3-18x-x^2+9)$
$=-18x^3-46x^2-8x+16$
2.
a.
$301^2=(300+1)^2=300^2+2.300+1=90000+600+1$
$=90601$
b.
$198^2=(200-2)^2=4(100-1)^2=4(100^2-2.100+1)$
$=4(10000-200+1)=4.9801=39204$
c.
$93.107=(100-7)(100+7)=100^2-7^2$
$=10000-49=9951$
d.
$127^2+146.127+73^2$
$=127^2+2.73.127+73^2$
$=(127+73)^2=200^2=40000$
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=3^{32}-1\)
\(\Leftrightarrow A=3^{31}-\frac{1}{2}\)