Question

Đã trả lời bởi chuyên gia

Answer 1

b,

${(x - 2)}^{2} - 4 = 0 \Leftrightarrow {(x - 2)}^{2} - 2^{2} = 0 \Leftrightarrow (x - 2 - 2) (x - 2 + 2) = 0 \Leftrightarrow (x - 4) . x = 0 \Leftrightarrow x - 4 = 0 h o ặ c x = 0 \Leftrightarrow x = 4 h o ặ c x = 0$

c,

$x^{3} - 19 x - 30 = 0 \Leftrightarrow x^{3} - 25 x + 6 x - 30 = 0 \Leftrightarrow x (x^{2} - 25) + 6 (x - 5) = 0 \Leftrightarrow x (x + 5) (x - 5) + 6 (x - 5) = 0 \Leftrightarrow (x - 5) (x^{2} + 5 x + 6) = 0 \Leftrightarrow (x - 5) (x^{2} + 2 x + 3 x + 6) = 0 \Leftrightarrow (x - 5) [x (x + 2) + 3 (x + 2)] = 0 \Leftrightarrow (x - 5) (x + 2) (x + 3) = 0 \Leftrightarrow x - 5 = 0 h o ặ c x + 2 = 0 h o ặ c x + 3 = 0 \Leftrightarrow x = 5 h o ặ c x = - 2 h o ặ c x = - 3$

d,

$2 x (x - 5) - x (3 + 2 x) = 0 \Leftrightarrow x (2 x - 10 - 3 - 2 x) = 0 \Leftrightarrow x . (- 13) = 0 \Leftrightarrow x = 0$

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Answer 2

b,

${(x - 2)}^{2} - 4 = 0 \Leftrightarrow {(x - 2)}^{2} - 2^{2} = 0 \Leftrightarrow (x - 2 - 2) (x - 2 + 2) = 0 \Leftrightarrow (x - 4) . x = 0 \Leftrightarrow x - 4 = 0 h o ặ c x = 0 \Leftrightarrow x = 4 h o ặ c x = 0$

c,

$x^{3} - 19 x - 30 = 0 \Leftrightarrow x^{3} - 25 x + 6 x - 30 = 0 \Leftrightarrow x (x^{2} - 25) + 6 (x - 5) = 0 \Leftrightarrow x (x + 5) (x - 5) + 6 (x - 5) = 0 \Leftrightarrow (x - 5) (x^{2} + 5 x + 6) = 0 \Leftrightarrow (x - 5) (x^{2} + 2 x + 3 x + 6) = 0 \Leftrightarrow (x - 5) [x (x + 2) + 3 (x + 2)] = 0 \Leftrightarrow (x - 5) (x + 2) (x + 3) = 0 \Leftrightarrow x - 5 = 0 h o ặ c x + 2 = 0 h o ặ c x + 3 = 0 \Leftrightarrow x = 5 h o ặ c x = - 2 h o ặ c x = - 3$

d,

$2 x (x - 5) - x (3 + 2 x) = 0 \Leftrightarrow x (2 x - 10 - 3 - 2 x) = 0 \Leftrightarrow x . (- 13) = 0 \Leftrightarrow x = 0$

Answer 3

`b)`

$(x-2)^2-4=0$

$⇔(x-2)^2-2^2=0$

$⇔(x-2-2).(x-2+2)=0$

$⇔(x-4).x=0$

$⇔$ $\left[ \begin{array}{l}x-4=0\\x=0\end{array} \right.$ `<=>` $\left[ \begin{array}{l}x=4\\x=0\end{array} \right.$

Vậy `x ∈ {4;0}`

`c)`

$x^3-19x-30=0$

$⇔ x^3-10x-9x-30=0$

$⇔ x.(x^2-9)-10.(x+3)=0$

$⇔ x.(x-3).(x+3)-10.(x+3)=0$

$⇔ (x+3).(x^2-3x-10)=0$

$⇔ (x+3).(x^2-5x+2x-10)=0$

$⇔ (x+3).[x.(x-5)+2.(x-5)]=0$

$⇔ (x+3).(x+2).(x-5)=0$

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

Vậy `x ∈ {-3;-2;5}`

`d)`

$2x.(x-5)-x.(3+2x)=0$

$⇔2x^2-10x-3x-2x^2=0$

$⇔ -13x=0$

$⇔x=0$

Vậy `x=0`

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