`@Kem`
Bài `6:`
`a, (3/4)² . (3/4)³ = (3/4)^(2+3) = (3/4)^5 = 243/1024`
`b, ((-1)/2 - 1/3)² = 1/4 - 1/9 = 5/36`
`c, ( 0,5 . 4/3)³ = 1/8 . 64/27 = 8/27`
`d, 2² - (5/7)^0 + (1/3)^4 . 3^6`
`= 4 - 1 + (3^6)/(3^4)`
`= 4 - 1 + 3^2`
`= 3 + 9`
`= 12`
`e, (9² . 3³)/(3^7) . 2020`
`= (3^4 . 3³)/(3^7) . 2020`
`= (3^7)/(3^7) . 2020`
`= 1 . 2020`
`= 2020`
Bài `7:`
`a, x² = (1/2)² ⇒ x = 1/2` hoặc `x = -1/2`
`b, (x-1)³ = 1/8`
`⇒ TH1: x- 1= 1/2 ⇔ x = 3/2`
`⇒ TH2: x - 1 = -1/2 ⇔ -1/2`
`c, (2x - 0,75)^4 = 16`
`⇒ TH1: 2x - 0,75 = 4 ⇔ 2x = 4,75 ⇔ x = 19/8`
`⇒ TH2: 2x - 0,75 = -4 ⇔ 2x = 3,25 ⇔ x = 13/8`
