Giải các phương trình sau $\frac{x-3}{5}=6-\frac{1-2x}{3}$

$$\begin{gathered} \frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5 \\ \iff \frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x+2}{7}-5 \end{gathered}$$

⇔ 14(5x – 3) – 21(7x – 1) = 12(4x + 2) – 5.84

⇔ 70x – 42 – 147x + 21 = 48x + 24 – 420

⇔ 70x – 147x – 48x = 24 – 420 + 42 – 21

⇔ -125x = -375

⇔ x = 3

Phương trình có nghiệm x = 3

$$\begin{gathered} \frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10} \\ \iff \frac{6x+3}{4}-5=\frac{6x-2}{5}-\frac{3x+2}{10} \\ \iff \frac{5\left(6x+3\right)-5.20}{20}=\frac{4\left(6x-2\right)-2\left(3x+2\right)}{20} \end{gathered}$$

⇔ 5. (6x +3) – 5.20 = 4( 6x – 2) – 2( 3x + 2)

⇔ 30x + 15 – 100 = 24x -8 – 6x -4

⇔ 30x – 85= 18x – 12

⇔ 30x – 18x = - 12 + 85

⇔ 12 x = 73

⇔ x = $\frac{73}{12}$

Vậy phương trình có nghiệm x = $\frac{73}{12}$