## giúp mk chuyên gia toán ơi

Question

giúp mk chuyên gia toán ơi

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4 years 2020-10-21T07:48:18+00:00 2 Answers 76 views 0

1. 8a) $x^2 + 2x – 24 = 0$

$\Leftrightarrow (x+6)(x -4) = 0$

$\Leftrightarrow \left[\begin{array}{l}x = -6\\x = 4\end{array}\right.$

b) $2x^2 + 5x – 3 = 0$

$\Leftrightarrow (x +1)\left(x -\dfrac{3}{2}\right) = 0$

$\Leftrightarrow \left[\begin{array}{l}x = -1\\x = \dfrac{3}{2}\end{array}\right.$

c) $3x^2 = -5x + 2$

$\Leftrightarrow (x +1)\left(x -\dfrac{2}{3}\right) = 0$

$\Leftrightarrow \left[\begin{array}{l}x = -1\\x = \dfrac{2}{3}\end{array}\right.$

9) $(x^2 +3x – 10)(x^2 – 5x – 6)$

$= (x + 5)(x -2)(x +1)(x – 6)$

$= [(x + 5)(x – 6)][(x -2)(x +1)]$

$= (x^2 – x – 30)(x^2 – x -2)$

2. Đáp án:

Giải thích các bước giải:

a) x^2+2x-24=0

⇔ x^2-4x+6x-24=0

⇔ x(x-4)+6(x-4)=0

⇔ (x+6)(x-4)=0

⇔ $$\left[ \begin{array}{l}x+6=0\\x-4=0\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}x=-6\\x=4\end{array} \right.$$

Vậy S={-6;4}

b) 2x^2+5x-3=0

⇔ 2x^2+6x-x-3=0

⇔ (2x-1)(x+3)=0

⇔ $$\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-3\end{array} \right.$$

Vậy S={1/2;-3}

c) 3x^2=-5x+2

⇔ 3x^2+5x-2=0

⇔ (3x-1)(x+2)=0

⇔ $$\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-2\end{array} \right.$$

Vậy S={1/3;-2}

9/

(x^2+3x-10)(x^2-5x-6)=(x^2-x-2)(x^2-x-30)

VT=x^4-2x^3-31x^2+32x+60

VP=x^4-2x^3-31x^2+32x+60

⇒ VT=VP

Vậy đẳng thức đã được chứng minh