9⁶⁴ - 1 = (9³² + 1)(9³² - 1) = ... = (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1)(9 + 1)(9 - 1) > (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9²​ + 1)(9 + 1)

9⁶⁴=(9³²​+1).(9³²-1)

=(9³²+1)(9¹⁶+1)(9¹⁶-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁸-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9⁴-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9²-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)(9-1)

Vì (9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)(9-1)>(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)

=>9⁶⁴-1>(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)

\(9^{64}-1\)

\(=\left(9^{32}+1\right)\left(9^{32}-1\right)\)

\(=\left(9^{32}+1\right)\left(9^{16}+1\right)\left(9^{16}-1\right)\)

\(...\)

\(=\left(9^{32}+1\right)\left(9^{16}+1\right)\left(9^8+1\right)\left(9^4+1\right)\left(9^2+1\right)\left(9+1\right)\left(9-1\right)\)

\(\Rightarrow9^{64}-1>\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

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

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

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

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

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

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

Vậy \(B< A\)

Ta có:

\(\left(-16\right)^{11}=-16^{11}=-\left(2^4\right)^{11}=-2^{4\cdot11}=-2^{44}\)

\(\left(-32\right)^9=-32^9=-\left(2^5\right)^9=-2^{5\cdot9}=-2^{45}\)

Mà: \(44< 45\)

\(\Rightarrow2^{44}< 2^{45}\)

\(\Rightarrow-2^{44}>-2^{45}\)

(−16)11=−1611=−(24)11=−24⋅11=−244

(−32)9=−329=−(25)9=−25⋅9=−245(−32)9=−329=−(25)9=−25⋅9=−245

Mà                 44<4544<45

⇒244<245⇒244<245

⇒−244>−245⇒−2            44>−245

b) \(9^5=3^{2\cdot5}=3^{10}\)

\(27^3=3^{3\cdot3}=3^9\)

=> tự kết luận

c) \(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2}^3\right)^6=\left(\frac{1}{2}\right)^{18}\)

\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2}^5\right)^4=\left(\frac{1}{2}\right)^{20}\)

=> tự kết luận

b) Ta có: \(9^5=\left(3^2\right)^5=3^{10}\) 

             \(27^3=\left(3^3\right)^3=3^9\)

Vì 10 > 9 => 3¹⁰ > 3⁹

Vậy 9⁵ > 27³

1. So sánh : 

b) 9^5 và 27^3 

9^5 = ( 3^2 )^5 = 3^10

27^3 = ( 3^3 )^3  = 3^9 

Vì 3^10 > 3^9 => 9^5 > 27^3 

Vậy 9^5 > 27^3 

c) \(\left(\frac{1}{8}\right)^6\)và \(\left(\frac{1}{32}\right)^4\)

\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2}\right)^{3.6}=\left(\frac{1}{2}\right)^{18}\)

\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2}\right)^{5.4}=\left(\frac{1}{2}\right)^{20}\)

Vì ( 1/2)^18 < (1/2)^20 => (1/8)^6 < (1/32)^4 

Vậy (1/8)^6 < (1/32)^4

a)  ta có \(\hept{\begin{cases}\left(-2\right)^{14}=\left(-2^7\right)^2=128^2\\5^6=5^{3.2}=\left(5^3\right)^2=125^2\end{cases}}\)

vì \(128>125\Rightarrow128^2>125^2\Rightarrow5^6< \left(-2\right)^{14}\)

vậy \(5^6< \left(-2\right)^{14}\)

b) ta có \(\hept{\begin{cases}27^3=\left(3^3\right)^3=3^9\\9^5=\left(3^2\right)^5=3^{10}\end{cases}}\)

vì \(3^{10}>3^9\Rightarrow9^5>27^3\)

vậy \(9^5>27^3\)

c) ta có \(\hept{\begin{cases}\left(\frac{1}{8}\right)^6=\left(\frac{1}{2}\right)^{18}\\\left(\frac{1}{32}\right)^4=\left(\frac{1}{2}\right)^{20}\end{cases}}\)

vì \(18< 20\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)

vậy \(\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)

Ta có: \(32^{27}=\left(2^5\right)^{27}=2^{135}\)

\(16^{39}=\left(2^4\right)^{39}=2^{156}\)

mà \(2^{135}< 2^{156}\)

nên \(32^{27}< 16^{39}\)

mà \(16^{39}< 18^{39}\)

nên \(32^{27}< 18^{39}\)

\(\Leftrightarrow-32^{27}>-18^{39}\)

\(\Leftrightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)

\(A=\left[0,8\cdot7+(0,8)^2\right]\cdot\left[1,25\cdot7-\frac{4}{5}\cdot1,25\right]-47,86\)

\(=0,8\cdot(7+0,8)\cdot1,25\cdot(7-0,8)-47,86\)

\(=0,8\cdot7,8\cdot1,25\cdot6,2-47,86\)

\(=48,36-47,86=0,5\)

\(B=\frac{(1,09-0,29)\cdot\frac{5}{4}}{(18,9-16,65)\cdot\frac{8}{9}}=\frac{0,8\cdot1,25}{2,25\cdot\frac{8}{9}}=\frac{1}{2}\)

\(A:B=0,5:\frac{1}{2}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}\cdot2=1\)

A gấp 1 lần B

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\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

                \(.........\)

\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)

Vậy  \(B< A\)

 A=1.853020189*10 \(^{15}\)

B= 9.265100944*10\(^{15}\)

tự so sánh

Xét B ta có:

\(2B=2\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^{16}-1\right).\left(3^{16}+1\right)\)

\(2B=3^{32}-1\)

\(B=\frac{3^{32}-1}{2}< A=3^{32}-1\)

Vậy B < A

Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)