giúp mk chuyên gia toán ơi Question giúp mk chuyên gia toán ơi in progress 0 Môn Toán Cherry 4 years 2020-10-21T07:48:18+00:00 2020-10-21T07:48:18+00:00 2 Answers 79 views 0
Answers ( )
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) $(x^2 +3x – 10)(x^2 – 5x – 6)$
$= (x + 5)(x -2)(x +1)(x – 6)$
$= [(x + 5)(x – 6)][(x -2)(x +1)]$
$= (x^2 – x – 30)(x^2 – x -2)$
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`
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