Đáp án:

a) `S={2/5;1/6}`

b) `S={-3;5/2}`

c) `S={2004;1/4}`

d) `S={-1;0}`

e) `S={3;-6}`

Giải thích các bước giải:

a) `6x(5x-2)-5x+2=0`

`<=>6x(5x-2)-(5x-2)=0`

`<=>(5x-2)(6x-1)=0`

`<=>`\(\left[ \begin{array}{l}5x-2=0\\6x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}5x=2\\6x=1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{2}{5}\\x=\dfrac{1}{6}\end{array} \right.\) 

Vậy: `S={2/5;1/6}`

b) `5(x+3)-2x(3+x)=0`

`<=>(x+3)(5-2x)=0`

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

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

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

Vậy: `S={-3;5/2}`

c) `4x(x-2004)-x+2004=0`

`<=>4x(x-2004)-(x-2004)=0`

`<=>(x-2004)(4x-1)=0`

`<=>`\(\left[ \begin{array}{l}x-2004=0\\4x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2004\\4x=1\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2004\\x=\dfrac{1}{4}\end{array} \right.\) 

Vậy: `S={2004;1/4}`

d) `(x+1)^2=x+1`

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

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

`<=>(x+1)x=0`

`<=>`\(\left[ \begin{array}{l}x+1=0\\x=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-1\\x=0\end{array} \right.\) 

Vậy: `S={-1;0}`

e) Sửa đề: `x^2+3x-18=0`

`<=>x^2-3x+6x-18=0`

`<=>x(x-3)+6(x-3)=0`

`<=>(x-3)(x+6)=0`

`<=>`\(\left[ \begin{array}{l}x-3=0\\x+6=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=3\\x=-6\end{array} \right.\) 

Vậy: `S={3;-6}`

Hướng dẫn trả lời:

Câu 1:

a) `6x*(5x - 2) - 5x + 2 = 0`

`↔ 6x*(5x - 2) - (5x - 2) = 0`

`↔ (6x - 1)*(5x - 2) = 0`

`↔ [(6x - 1 = 0), (5x - 2 = 0):}`

`↔ [(6x = 1), (5x = 2):}`

`↔ [(x = 1/6), (x = 2/5):}`

Vậy `x = 1/6` hoặc `x = 2/5`

b) `5*(x + 3) - 2x*(3 + x) = 0`

`↔ 5*(x + 3) - 2x*(x + 3) = 0`

`↔ (5 - 2x)*(x + 3) = 0`

`↔ [(5 - 2x = 0), (x + 3 = 0):}`

`↔ [(2x = 5), (x = - 3):}`

`↔ [(x = 5/2), (x = - 3):}`

Vậy `x = 5/2` hoặc `x = -3`

c) `4x*(x - 2004) - x + 2004 = 0`

`↔ 4x*(x - 2004) - (x - 2004) = 0`

`↔ (4x - 1)*(x - 2004) = 0`

`↔ [(4x - 1 = 0), (x - 2004 = 0):}`

`↔ [(4x = 1), (x = 2004):}`

`↔ [(x = 1/4), (x = 2004):}`

Vậy `x = 1/4` hoặc `x = 2004`

d) `(x + 1)^2 = x + 1`

`↔ (x + 1)^2 - (x + 1) = 0`

`↔ (x + 1)*[(x + 1) - 1] = 0`

`↔ (x + 1)*(x + 1 - 1) = 0`

`↔ x*(x + 1) = 0`

`↔ [(x = 0), (x + 1 = 0):}`

`↔ [(x = 0), (x = - 1):}`

Vậy `x = 0` hoặc `x = -1`

e) Đề sai nên mình sửa nhé!

- Đề sửa: `x^2 + 3x - 18 = 0`

`↔ x^2 - 3x + 6x - 18 = 0`

`↔ (x^2 - 3x) + (6x - 18) = 0`

`↔ x*(x - 3) + 6*(x - 3) = 0`

`↔ (x + 6)*(x - 3) = 0`

`↔ [(x + 6 = 0), (x - 3 = 0):}`

`↔ [(x = - 6), (x = 3):}`

Vậy `x = -6` hoặc `x = 3`