Câu 1: Giải các phương trình sau: a, 1/3x-3=6x+1 b,(x+5)(2x+3)+x^2-25=0 c,2/x+1+4/x-2=3x/x^2-x-2

Đáp án: + Giải thích các bước giải:

`a//`

`1/(3x-3) = 6x + 1`         `ĐKXĐ : x \ne 1`

`⇔ ( 3x – 3 )( 6x + 1 ) = 1`

`⇔ 18x^2 – 15x – 3 =1`

`⇔ 18x^2 – 15x – 4 = 0`

`⇔ x = (-b±\sqrt{b^2-4ac})/(2a)`

`⇔ a = 18 ; b = -15 ; c = -4`

`⇔ x = (-(-15)±\sqrt{(-15)^2 – 4.18.(-4)))/2.18`

`⇔ x = (5 + sqrt{57})/12 ; (5 – sqrt{57})/12`

Vậy `S = { (5 ± sqrt{57})/12 }`

`b//`

`(x+5)(2x+3)+x^2-25=0`

`⇔ 2x^2 + 3x + 10x + 15 + x^2 – 25 = 0`

`⇔ 2x^2 + 13 – 10 + x^2 = 0`

`⇔ 3x^2 + 13x – 10 = 0`

`⇔ 3x^2 + 15x – 2x – 10 = 0`

`⇔ (3x – 2)(x + 5) =0`

`⇔` \(\left[ \begin{array}{l}3x=2\\x=0-5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=-5\end{array} \right.\) 

Vậy `S = { 2/3 ; -5 }`

`c//`

`2/(x+1) + 4/(x-2) = (3x)/(x^2-x-2)`            `ĐKXĐ : x \ne -1 ; 2`

`⇔ (2(x – 2)(x+1))/((x+1)(x-2)) + (4(x+1)(x-2))/((x-2)(x+1)) = (3(x-2))/((x+1)(x-2)`

`⇔ 2(x+1) + 4(x-2) = 3(x-2)`

`⇔ 2x + 2 + 4x – 8 = 3x – 6`

`⇔ 2x – 6 + 4x = 3x – 6`

`⇔ 6x – 6 = 3x – 6`

`⇔ 6x = 3x`

`⇔ 3x = 0`

`⇔ x = 0`

Vậy `S = { 0 }`