a ) 2|x - 3| - 5 = 3 <=> 2|x - 3| = 8 <=> |x - 3| = 4 => x - 3 = ± 4

TH1 : x - 3 = 4 => x = 7

TH2 : x - 3 = - 4 => x = - 1

Vậy x = { - 1; 7 }

b ) 2|2x + 3| + |2x + 3| = 6 <=> 3|2x + 3| = 6 => |2x + 3| = 2 => 2x + 3 = ± 2

=> x = { - 5/2 ; - 1/2 }

c ) 3|x + 1|² + |x + 1|² = 16

4|x + 1|² = 16

=> |x + 1|² = 4 = 2² ( ko xét TH |x + 1| = - 2 vì |x + 1| ≥ 0 )

=> |x + 1| = 2 => x + 1 = ± 2 => x = { - 3; 1 }

\(\frac{2}{3}\)

| x - 3 | = | 2x + 1 |

\(\Rightarrow\orbr{\begin{cases}x-3=2x+1\\x-3=-\left(2x+1\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-2x=1+3\\x+2x=-1+3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=4\\3x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=\frac{2}{3}\end{cases}}\)

I x - 3 I = I 2x + 1 I   ₍₁₎

Xét \(x< \frac{-1}{2}\)ta co   \(\left(1\right)\Leftrightarrow3-x=-2x-1\Leftrightarrow x=-4\)(thỏa mãn)

Xét  \(\frac{-1}{2}\le x< 3\)ta có \(\left(1\right)\Leftrightarrow3-x=2x+1\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)(thỏa mãn)

Xét \(x\ge3\)ta có \(\left(1\right)\Leftrightarrow x-3=2x+1\Leftrightarrow x=-4\)(loại vì \(x\ge3\))

I2x+1/3I+3/4=0

|2x+1/3|=-3/4

TH1:

2x+1/3=-3/4

2x=-3/4-1/3

2x=-13/12

x=-13/12:2

x=-13/24

TH2:

2x+1/3=3/4

2x=3/4-2/3

2x=5/12

x=5/12:2

x=5/24

vậy x=-13/24;5/24