~ Gửi bạn nhea ~
`a//`
Ta có : `x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)`
Mà `(2+x)(x+2) = x^2 - 4`
ĐKXĐ : `x^2 - 4 \ne 0 ⇔ x^2 \ne 4 ⇔ |x| \ne 2 ⇔ x \ne \pm2`
`A = (x/(x^2-4)+2/(2-x)+1/(x+2))*(x+2)/2`
`= (x/((x-2)(x+2))-2/(x-2)+1/(+2)) * (x+2)/2`
`= (x-2(x+2)+x-2)/((x-2)(x+2)) * (x+2)/2`
`= (x-2x-4+x-2)/((x-2)(x+2)) * (x+2)/2`
`= (-6)/((x-2)(x+2)) * (x+2)/2`
`= (-3)/(x-2)`
`b//`
Giá trị `A` khi `x = -1` :
`A = (-3)/(-1-2) = (-3)/(-3) = 1`
`2//`
`a//`
`x(x-4) + 1 = 3x - 5`
`⇔ x^2 - 4x + 1 = 3x - 5`
`⇔ x^2 - 4x + 1 - 3x + 5 = 0`
`⇔ x^2 - 7x + 6 = 0`
`⇔ (x^2-x) + (-6x+6) = 0`
`⇔ x(x-1) - 6(x-1) = 0`
`⇔ (x-1)(x-6) = 0`
TH1 : `x - 1 = 0 <=> x = 1`
TH2 : `x - 6 = 0 <=> x = 6`
Vậy `S = {1,6}`
`b//`
`2x^3 - 3x^2 - 2x + 3 = 0`
`⇔ (2x^3-3x^2) + (-2x+3) = 0`
`⇔ -(2x-3) + x^2(2x-1) = 0`
`⇔ (2x-3)(x^2-1) = 0`
`⇔ (2x-3)(x-1)(x+1) = 0`
TH1 : `2x - 3 = 0 <=> 2x = 3 <=> x = 3/2`
TH2 : `x - 1 = 0 <=> x =1`
TH3 : `x + 1 = 0 <=> x = -1`
Vậy `S = {-1,1,3/2}`