`7,`

`(3x-4)^2 - (x+2)^2 = 0`

`<=> (3x-4-x-2).(3x-4+x+2) = 0`

`<=> (x-6).(4x-2) = 0`

`TH1`

`x-6 = 0`

`x = 6`

TH2`

`4x-2 = 0`

`x = 1/2`

Vậy `x = {6 ; 1/2}`

`8,`

`(2x-1)^2 - (3-x)^2 = 0`

`<=> (2x-1-3+x).(2x-1+3-x) = 0`

`<=> (3x-4).(x + 2) = 0`

`TH1`

`3x-4 = 0`

`x = 4/3`

`TH2`

`x+2 = 0`

`x = -2`

Vậy `x = {4/3 ; -2}`

`9,`

`(5x-1)^2 - (x+1)^2 = 0`

`<=> (5x-1-x-1).(5x-1+x+1) = 0`

`<=> (4x-2).6x = 0`

`TH1`

`4x-2 = 0`

`x = 1/2`

`TH2`

`6x = 0`

`x = 0`

Vậy `x = {1/2 ; 0}`

#Yoshimoto Shizuka