Question

Bài 2: Giải các phương trình đưa về dạng tích sau:
a) (3x – 2)(4x + 5) = 0 b) x(2x – 3 ) – 4x + 6 =0
c) 3x(x – 1) + 2(x – 1 ) =0 d) x 2 – 1 = x(x – 1)

Answer 1

a)(3x-2)(4x-5)=0

3x-2=0 hoặc 4x-5=0

x=\(\frac{2}{3}\) hoặc x=\(\frac{5}{4}\)

b)x(2x-3)-4x+6=0

x(2x-3)-2(2x-3)=0

(2x-3)(x-2)=0

2x-3=0 hoặc x-2=0

x=\(\frac{3}{2}\) hoặc x=2

c)3x(x-1)+2(x-1)=0

(x-1)(3x+2)=0

x-1=0 hoặc 3x+2=0

x=1 hoặc x=\(\frac{-2}{3}\)

d)x²-1=x(x-1)

(x+1)(x-1)=x(x-1)

(x+1)(x-1)-x(x-1)=0

(x-1)(x+1-x)=0

(x-1)1=0

x-1=0

x=1

Answer 2

a) (3x - 2)(4x + 5) = 0

\(\Leftrightarrow\) 3x - 2 = 0 hoặc 4x + 5 = 0

* 3x - 2 = 0 * 4x + 5 = 0

\(\Rightarrow\) 3x = 2 \(\Rightarrow\) 4x = -5

\(\Rightarrow\) x = \(\frac{2}{3}\) \(\Rightarrow x=-\frac{5}{4}\)

vậy x \(\in\) \(\left\{\frac{2}{3};-\frac{5}{4}\right\}\)

b) x (2x - 3) - 4x + 6 = 0

\(\Rightarrow x\left(2x-3\right)-2\left(2x-3\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(x-2\right)=0\)

\(\Rightarrow2x-3=0\) hoặc \(\left(x-2\right)=0\)

* \(2x-3=0\) * \(x-2-0\)

\(\Rightarrow2x=3\) \(\Rightarrow x=2\)

\(\Rightarrow x=\frac{3}{2}\)

vậy \(x\in\left\{\frac{3}{2};2\right\}\)

Answer 3

c) \(3x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Rightarrow\left(3x+2\right)\left(x-1\right)=0\)

\(\Rightarrow3x+2=0\) hoặc \(x-1=0\)

* \(3x+2=0\) * \(x-1=0\)

\(\Rightarrow3x=-2\) \(\Rightarrow x=1\)

\(\Rightarrow x=-\frac{2}{3}\)

vậy pt trên có \(x\in\left\{-\frac{2}{3};1\right\}\)