giải pt `a,(2x+5)(x-4)+3x-12=0` `b, (2-1)^2 -(x+1)^2=0` `c, x^3-2x^2+2-x=0` `d, x^2+10x+25=(x-1)(x+5)`

Đáp án:

`a,` `(2x+5).(x−4)+3x−12=0`

`⇔ (2x+5).(x−4)+3.(x - 4) = 0`

`⇔ (x-  4).(2x + 5 + 3) = 0`

`⇔ (x-  4).(2x + 8) = 0`

`⇔ 2.(x - 4).(x + 4) = 0`

`⇔ (x - 4).(x + 4) = 0`

`⇔` \(\left[ \begin{array}{l}x - 4 = 0\\x + 4 = 0\end{array} \right.\)

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

Vậy `S = {4;-4}`

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`b,` `(2−1)^2−(x+1)^2=0`

`⇔ 1^2 - (x+1)^2=0`

`⇔ (x+1)^2 = 1`

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

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

Vậy `S = {0; -2}`

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`c,` `x^3−2x^2+2−x=0`

`⇔x^3−2x^2−x+2=0`

`⇔x^2.(x−2)−(x−2)=0`

`⇔(x−2).(x^2−1)=0`

`⇔(x−2).(x−1).(x+1)=0`

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

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

Vậy `S = {+-1; 2}`

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`d,` `x^2+10x+25=(x−1).(x+5)`

`⇔(x+5)^2−(x−1).(x+5)=0`

`⇔(x+5).(x+5−x+1)=0`

`⇔6.(x+5)=0`

`<=> x + 5 = 0`

`<=> x = -5`

Vậy `S = {-5}`

`#dariana`