:а)3^2x-1-3^2x+3^2x+3=237 b)5^2x-30*5^x+125=0 - вопрос №3213232323752305125013610257299496 от melochek228 18.02.2020 11:00

Question

Алгебра: :а)3^2x-1-3^2x+3^2x+3=237 b)5^2x-30*5^x+125=0 c) (1/36)^-10 корень x=2^5x*3^5x

melochek228 · Answer

3^( 2x - 1 ) - 3^2x + 3^( 2x + 3 ) = 237
3^2x * ( ( 1/3 ) - 1 + 27 ) = 237
3^2x = 237 : 26 1/3
3^2x = 237 : 79/3
3^2x = 9
3^2x = 3^2
2x = 2
x = 1

5^( 2x - 30 ) - 30*5^( x + 125 ) = 0
5^( 2x - 30 ) = 30*5^( x + 125 )
5 ^ ( 2x - 30 ) : 5 ^ ( x + 125 ) = 30
5 ^ ( 2x - 30 - x - 125 ) = 30
5 ^ ( x - 155 ) = 30
x - 155 = log₅ 30
x = ( log ₅ 30 ) + 155

( 1/36 ) ^ - 10√x = 2^5x * 3^ 5x
( 6 ^ - 2 ) ^ - 10√x = 6 ^ 5x
6 ^ 20√x = 6 ^ 5x
20√x = 5x
√x = a
20a = 5a^2
20a - 5a^2 = 0
5a( 4 - a ) = 0
5a = 0 ==> a = 0
4 - a = 0 ==> a = 4
√ x = 0 ==> x = 0
√ x = 4 ; x = 16