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)

\(9^{64}-1\)

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

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

\(...\)

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

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

chịu

\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)

\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)

\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)

\(=\frac{6}{2011}\)

Ta có A = 2018.2020 + 2019.2021

= (2020 - 2).2020 + 2019.(2019 + 2) 

= 2020² - 2.2020 + 2019² + 2.2019

= 2020² + 2019² - 2(2020 - 2019) = 2020² + 2019² - 2 = B

=> A = B

b) Ta có B = 9⁶⁴ - 1= (9³²)² - 1² 

= (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).80 

mà A =   (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1).10

=> A < B

c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)

=> A < B

d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)

=> A < B

a) \(85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2\)

\(=\left(85^2-15^2\right)+\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\)

\(=\left(85-15\right)\left(85+15\right)+\left(75-25\right)\left(75+25\right)+\left(65-35\right)\left(65+35\right)+\left(55-45\right)\left(55+45\right)\)

\(=70.100+50.100+30.100+10.100\)

\(=7000+5000+3000+1000\)

\(=16000\)

b) \(\frac{135^2+130.135+65^2}{135^2-65^2}\)

\(=\frac{135^2+2.60.135+65^2}{135^2-65^2}\)

\(=\frac{\left(135+65\right)^2}{\left(135-65\right)^2}\)

\(=\frac{200^2}{70^2}\) \(=\frac{200}{70}=\frac{20}{7}\)

\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)

\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)