Question

Giải các phương trình sau:

1. a) 7x + 12 = 0 b) 5x – 2 = 0 c) 12 – 6x = 0 d) – 2x + 14 = 0

2. a) 3x + 1 = 7x – 11 b) 2x + x + 12 = 0 c) x – 5 = 3 – x d) 7 – 3x = 9 – x

e) 5 – 3x = 6x + 7 f) 11 – 2x = x – 1 g) 15 – 8x = 9 – 5

3. a) 0,25x + 1,5 = 0 b) 6,36 – 5,2x = 0

Answer 1

Bài 1:

a) Ta có: 7x+12=0

\(\Leftrightarrow7x=-12\)

hay \(x=-\frac{12}{7}\)

Vậy: \(x=-\frac{12}{7}\)

b) Ta có: 5x-2=0

\(\Leftrightarrow5x=2\)

hay \(x=\frac{2}{5}\)

Vậy: \(x=\frac{2}{5}\)

c) Ta có: 12-6x=0

\(\Leftrightarrow6x=12\)

hay x=2

Vậy: x=2

d) Ta có: -2x+14=0

⇔-2x=-14

hay x=7

Vậy: x=7

Bài 2:

a) Ta có: 3x+1=7x-11

⇔3x+1-7x+11=0

⇔-4x+12=0

⇔-4x=-12

hay x=3

Vậy: x=3

b) Ta có: 2x+x+12=0

⇔3x+12=0

⇔3x=-12

hay x=-4

Vậy: x=-4

c) Ta có: x-5=3-x

⇔x-5-3+x=0

⇔2x-8=0

⇔2x=8

hay x=4

Vậy: x=4

d) Ta có: 7-3x=9-x

⇔7-3x-9+x=0

⇔-2x-2=0

⇔-2x=2

hay x=-1

Vậy: x=-1

e) Ta có: 5-3x=6x+7

⇔5-3x-6x-7=0

⇔-9x-2=0

⇔-9x=2

hay \(x=\frac{-2}{9}\)

Vậy: \(x=\frac{-2}{9}\)

f) Ta có: 11-2x=x-1

⇔11-2x-x+1=0

⇔12-3x=0

⇔3x=12

hay x=4

Vậy: x=4

g) Ta có: 15-8x=9-5

⇔15-8x=4

⇔8x=11

hay \(x=\frac{11}{8}\)

Vậy: \(x=\frac{11}{8}\)

Bài 3:

a) Ta có: 0,25x+1,5=0

⇔0,25x=-1,5

hay x=-6

Vậy: x=-6

b) Ta có: 6,36-5,2x=0

⇔5,2x=6,36

hay \(x=\frac{159}{130}\)

Vậy: \(x=\frac{159}{130}\)