Question

ccTìm x biết

a) \(3x^2\)-x-10=0

b) 5x+2 >_0

Answer 1

a) \(3x^2-x-10=0\)

<=> \(3x^2-6x+5x-10=0\)

<=> (\(3x^2-6x\))+(\(5x-10\)) = 0

<=> \(3x\left(x-2\right)\)+\(5\left(x-2\right)\) = 0

<=> \(\left(x-2\right)\left(3x+5\right)=0\)

<=> x - 2 =0 hoặc 3x +5 = 0

* x - 2 = 0 * 3x + 5 = 0

<=> x = 2 <=> 3x =-5

<=> x = \(\dfrac{-5}{3}\)

Vậy x = 2 hoặc x = \(\dfrac{-5}{3}\)

b) 5x +2 \(\ge\)0

<=>5x \(\ge\) -2

<=> x \(\ge\)\(\dfrac{-2}{5}\)

Vậy x \(\ge\)\(\dfrac{-2}{5}\)

Answer 2

a) 3x² - x - 10 = 0

⇔ 3x² + 5x - 6x - 10 = 0

⇔ x( 3x + 5) - 2( 3x + 5) = 0

⇔ ( x - 2)( 3x + 5) = 0

⇔ x = 2 hoặc : x = \(\dfrac{-5}{3}\)

KL....

b) 5x + 2 ≥ 0

⇔ x ≥ \(\dfrac{-2}{5}\)

KL....