Ta xét các trường hợp:

Trường hợp 1: \(x < \frac{1}{4}\)

• Khi đó \(x - 3 < 0 \Rightarrow \mid x - 3 \mid = 3 - x\).
• Đồng thời \(4 x - 1 < 0 \Rightarrow \mid 4 x - 1 \mid = 1 - 4 x\).

Suy ra:

\(A = 2 \left(\right. 3 - x \left.\right) - \left(\right. 1 - 4 x \left.\right) = 5 + 2 x .\)

Trường hợp 2: \(\frac{1}{4} \leq x < 3\)

• Khi đó \(x - 3 < 0 \Rightarrow \mid x - 3 \mid = 3 - x\).
• Đồng thời \(4 x - 1 \geq 0 \Rightarrow \mid 4 x - 1 \mid = 4 x - 1\).

Suy ra:

\(A = 2 \left(\right. 3 - x \left.\right) - \left(\right. 4 x - 1 \left.\right) = 7 - 6 x .\)

Trường hợp 3: \(x \geq 3\)

• Khi đó \(x - 3 \geq 0 \Rightarrow \mid x - 3 \mid = x - 3\).
• Đồng thời \(4 x - 1 \geq 0 \Rightarrow \mid 4 x - 1 \mid = 4 x - 1\).

Suy ra:

\(A = 2 \left(\right. x - 3 \left.\right) - \left(\right. 4 x - 1 \left.\right) = - 2 x - 5.\)

Kết luận:

\(A = \left{\right. 5 + 2 x & \text{n} \overset{ˊ}{\hat{\text{e}}} \text{u}\&\text{nbsp}; x < \frac{1}{4} , \\ 7 - 6 x & \text{n} \overset{ˊ}{\hat{\text{e}}} \text{u}\&\text{nbsp}; \frac{1}{4} \leq x < 3 , \\ - 2 x - 5 & \text{n} \overset{ˊ}{\hat{\text{e}}} \text{u}\&\text{nbsp}; x \geq 3.\)

2(x-3)-(4x-1)
= 2x-6 -(4x-1)
= 2x-6-4x+1
=-2x-5

Đặt A=2|x-3|-|4x-1|

=|2x-6|-|4x-1|

TH1: \(x<\frac14\)

=>4x-1<0; 2x-6<0

=>A=6-2x-(1-4x)

=6-2x-1+4x

=2x+5

TH2: \(\frac14\le x<3\)

=>4x-1>=0; 2x-6<0

=>A=-(4x-1)+(6-2x)

=-4x+1+6-2x

=-6x+7

TH3: x>=3

=>4x-1>0; 2x-6>=0

=>A=2x-6-(4x-1)

=2x-6-4x+1

=-2x-5