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\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=8.\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\\ =\left(3^8-1\right)....\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\cdot\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}-1\right)\\ ...\\ 2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\\ 2A=3^{128}-1\)
Vậy \(A=\dfrac{3^{128}-1}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các bước giải chi tiết
a) \(\dfrac{392-x}{32}\) + \(\dfrac{390-x}{34}\) + \(\dfrac{388-x}{36}\) = -3
⇔ \(\dfrac{392-x}{32}\)+1+\(\dfrac{390-x}{34}\)+1+\(\dfrac{388-x}{36}\)+1 = 0
⇔\(\dfrac{424-x}{32}\)+\(\dfrac{424-x}{34}\)+\(\dfrac{424-x}{36}\)=0
⇔\(\left(424-x\right)\)\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\right)\)=0
⇔\(424-x\) = 0\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\ne\forall x\right)\)
⇔\(x=424\)
b) \(\dfrac{x-3}{3}\)- \(x\) = \(5-\dfrac{x+1}{4}\)
⇔\(\dfrac{x-3-3x}{3}\) = 5 + \(\dfrac{-\left(x+1\right)}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = 5 + \(\dfrac{-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{20-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{-x+19}{4}\)
⇔ \(4\left(-2x-3\right)\) = \(3\left(-x+19\right)\)
⇔\(-8x-12\) = \(-3x+57\)
⇔\(-8x\) = \(-3x+69\)
⇔\(-5x=69\)
⇔ \(x=-\dfrac{69}{5}\)
(392-x)/32+(390-x)/34+(388-x)/36=-3
=>392-x)/32 +1 + (390-x)/34 +1 +(388-x)/36 +1=0
=>(424-x)/32+(424-x)/34+(424-x)/36=0
=>424-x=0(vì 1/32+1/34+1/36 khác 0)
=>x=424
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Khi x=3 thì \(A=\dfrac{3+2}{3-1}=\dfrac{5}{2}\)
b: \(B=\dfrac{x-1}{x}+\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{x^2-1+2x+1}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
\(P=A:B=\dfrac{x+2}{x-1}\cdot\dfrac{x+1}{x+2}=\dfrac{x+1}{x-1}\)
3: Để P>1/3 thì \(P-\dfrac{1}{3}>0\)
=>\(\Leftrightarrow3\left(x+1\right)-x+1>0\)
=>3x+3-x+1>0
=>2x+4>0
hay x>-2