Đáp án: + Giải thích các bước giải:
Bài 1 :
`a//`
`3x-20=5-2x`
`⇔ 3x + 2x = 20 + 5`
`⇔ 5x = 25`
`⇔ x = 5`
Vậy `S = {5}`
`b//`
`(2x-5)/(x+3) + 4/((1-x)(x+3)) = (2x)/(x-1)`
ĐKXĐ : `x \ne -3 , x \ne 1`
`⇔ (2x-5)(-x+1)(x-1)-4(-x+1)=2x(x+3)(-x+1)`
`⇔ -2x^3 + 13x^2 - 14x + 1 = -2x^3`
`⇔ 13x^2 - 14x + 1 =0`
`⇔ (x-1)(13x-1)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\13x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\text{(không thỏa mãn)}\\x=\dfrac{1}{13}\text{(thỏa mãn)}\end{array} \right.\)
Vậy `S = { 1/13 }`
`c//`
`x^2(x^2+2)-4x^2=8`
`⇔ x^4 - 2x^2 = 8`
`⇔ x^4 - 2x^2 - 8 =0`
`⇔ x = 2,-2,i\sqrt{2}.-i\sqrt{2}`
Bài 2 :
`a//`
`(x-3)(x-4)-(x-2)^2=1`
`⇔ -3x + 8 = 1`
`⇔ 3x = 7`
`⇔ x = 7/3`
Vậy `S = { 7/3}`
`b//`
`(x-3)/(2x-1) - (x+8)/(2x+1) = 25/(4x^2-1)`
`⇔ (x-3)(2x+1)-(x+8)(2x-1)=25`
`⇔ -20x + 5 = 25`
`⇔ -20x = 20`
`⇔ x = 1`
Vậy `S = {1}`
`c//`
`|2x-3|=3-2x`
Đk : `x < 3/2 , x ≥ 3/2`
`2x - 3 = 3 - 2x`
`⇔2x = -2x + 6`
`⇔4x = 6`
`⇔ x = 3/2\text{(thỏa mãn)}`
