Question

Bài 1. Giải các phương trình sau:

a. 3x - 6 + x = 9 - x

b. x(x-2) - 2x(x+3) = 21 - x² - x

c. \(\frac{2x-3}{4}\) + 2 = \(\frac{1-x}{6}\)

Answer 1

Bài 1:

a) Ta có: 3x-6+x=9-x

⇔4x-6-9+x=0

⇔5x-15=0

⇔5(x-3)=0

mà 5≠0

nên x-3=0

hay x=3

Vậy: x=3

b) Ta có: \(x\left(x-2\right)-2x\left(x+3\right)=21-x^2-x\)

\(\Leftrightarrow x^2-2x-2x^2-6x-21+x^2+x=0\)

\(\Leftrightarrow-7x-21=0\)

\(\Leftrightarrow-7\left(x+3\right)=0\)

mà -7≠0

nên x+3=0

hay x=-3

Vậy: x=-3

c) Ta có: \(\frac{2x-3}{4}+2=\frac{1-x}{6}\)

\(\Leftrightarrow\frac{3\left(2x-3\right)}{12}+\frac{24}{12}=\frac{2\left(1-x\right)}{12}\)

\(\Leftrightarrow3\left(2x-3\right)+24=2\left(1-x\right)\)

\(\Leftrightarrow6x-9+24=2-2x\)

\(\Leftrightarrow6x+15-2+2x=0\)

\(\Leftrightarrow8x+13=0\)

\(\Leftrightarrow8x=-13\)

hay \(x=\frac{-13}{8}\)

Vậy: \(x=\frac{-13}{8}\)