$#Aridoto$

`a, (4x-3)(5x+6)=0`

\(\left[ \begin{array}{l}4x-3=0\\5x+6=0\end{array} \right.\)

\(\left[ \begin{array}{l}4x=3\\5x=-6\end{array} \right.\) 

\(\left[ \begin{array}{l}x=\frac{3}{4}\\x=\frac{-6}{5}\end{array} \right.\)

Vậy `S={3/4;(-6)/5}`

`b, 1/(x-1)-3/(x-2)=(-1)/[(x-1)(x-2)]`   $ĐKXĐ: \left \{ {{x\ne1} \atop {x\ne2}} \right.$ 

`↔(x-2)/[(x-1)(x-2)]-[3(x-1)]/[(x-1)(x-2)]=(-1)/[(x-1)(x-2)]`

`↔x-2-3(x-1)=-1`

`↔x-2-3x+3=-1`

`↔x-3x=-1-3+2`

`↔-2x=-2`

`↔x=1` (k t/m đkxđ)

Vậy `S=∅`

`c, |3x|=x+2`

Ta có: `x+2>0↔x>-2`

\(\left[ \begin{array}{l}3x=x+2\\3x=-x-2\end{array} \right.\)

\(\left[ \begin{array}{l}3x-x=2\\3x+x=-2\end{array} \right.\) 

\(\left[ \begin{array}{l}2x=2\\4x=-2\end{array} \right.\) 

\(\left[ \begin{array}{l}x=1\\4x=-1/2\end{array} \right.\) 

Vậy `S={1;-1/2}`

`d, 2x-7=5x+20`

`↔2x-5x=20+7`

`↔-3x=27`

`↔x=27:(-3)`

`↔x=-9`

Vậy `S={-9}`

Đáp án+Giải thích các bước giải:

`a) (4x - 3)(5x + 6) = 0`

`<=> [(4x - 3 = 0),(5x + 6 = 0):}`

`<=> [(x = 3/4),(x = - 6/5):}`

Vậy `S = {3/4 ; - 6/5}`

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`b) 1/{x - 1} - 3/{x - 2} = {-1}/{(x - 1)(x - 2)}` (ĐKXĐ: `x ne 1 ; ne 2`)

`<=> {x  - 2}/{(x - 1)(x - 2)} - {3(x - 1)}/{(x - 1)(x - 2)} = {-1}/{(x - 1)(x - 2)}`

`=> x - 2 - 3(x - 1) = - 1`

`<=> x - 2 - 3x + 3 = -1`

`<=> - 2x + 1 = -1`

`<=> - 2x = - 2`

`<=> x = 1` (Loại)

Vậy `S = \emptyset`

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`c) |3x| = x + 2` (đk: `x + 2 >= 0 <=> x >= - 2`)

`<=> [(3x = x + 2),(3x = - x - 2):}`

`<=> [(3x - x = 2),(3x + x = -2):}`

`<=> [(2x = 2),(4x = - 2):}`

`<=> [(x = 1(Nhận)),(x = - 1/2(Nhận)):}`

Vậy `S = {1 ; -1/2}`

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`d) 2x - 7 = 5x + 20`

`<=> 2x - 5x = 7 + 20`

`<=> - 3x = 27`

`<=> x = - 9`

Vậy `S = {-9}`