Question

if 2019^x+2019^-x=3 then find the value of √2019^6x-2019^-x÷2019^x-2019^-x find related solutions​

Answer 1

Given : 2019^x+2019^-x=3

To Find : value of √2019^6x-2019^-x÷2019^x-2019^-x

Solution:

2019^x+2019^-x=3

Let say 2019^x = a

=> 2019^-x = 1/a

a + 1/a = 3

=> a² + 1 = 3a

(√2019^6x-2019^-x)÷ (2019^x-2019^-x)

=  (2019^3x - 2019^-x)÷ (2019^x-2019^-x)

= ( a³ - 1/a) ÷ (a -1/a)

multiplying numerator and denominator by a

= ( a⁴ - 1)  ÷ (a² - 1)

= (a² + 1)(a² - 1)  ÷ (a² - 1)

= a² + 1

= 3a

 a² + 1 = 3a

=> a² - 3a + 1 = 0

=> a  = (3 ± √9-4)/2

=> a = (3 ± √5)/2

√2019^6x-2019^-x÷2019^x-2019^-x =   3  (3 ± √5)/2

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Answer 2

Answer:

Answer:12

Answer:12Step-by-step explanation:

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-x

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:2019^x+2019^-x=3

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:2019^x+2019^-x=3Let say 2019^x = a => 2019^-x = 1/a

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:2019^x+2019^-x=3Let say 2019^x = a => 2019^-x = 1/aa + 1/a = 3

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:2019^x+2019^-x=3Let say 2019^x = a => 2019^-x = 1/aa + 1/a = 3=> a² + 1 = 3a

Answer:12Step-by-step explanation:Given: 2019^x+2019^-x=3To Find : value of√2019 6x-2019^-x÷2019^x-2019^-xSolution:2019^x+2019^-x=3Let say 2019^x = a => 2019^-x = 1/aa + 1/a = 3=> a² + 1 = 3a(√2019^6x-2019^-x) (2019^x-2019^-x)

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3a

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3aa²+1=3a

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3aa²+1=3a=> a²-3a + 1 = 0 => a =(3+√9-4)/2 => a = (3 ± √5)/2

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3aa²+1=3a=> a²-3a + 1 = 0 => a =(3+√9-4)/2 => a = (3 ± √5)/2therefore on solving it

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3aa²+1=3a=> a²-3a + 1 = 0 => a =(3+√9-4)/2 => a = (3 ± √5)/2therefore on solving it√2019^6x-2019^-x÷2019^x-2019^-x = 3 (3 ±

= (2019^3x - 2019^-x) (2019^x-2019^-x) =(a³-1/a) (a -1/a) multiplying numerator and denominator by a = (a¹ - 1) = (a² - 1) = (a² + 1)(a²-1) = (a² - 1)= a² + 1= 3aa²+1=3a=> a²-3a + 1 = 0 => a =(3+√9-4)/2 => a = (3 ± √5)/2therefore on solving it√2019^6x-2019^-x÷2019^x-2019^-x = 3 (3 ±√5)/2 =12