a, \(\left|2x-5\right|=4\)
\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}\)
b, \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Rightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
c, \(\left|x+3\right|-\left|3x+2\right|=x+2\)
Ta có: x + 3 = 0 => x = -3
3x + 2 = 0 => x = -2/3
Lập bảng xét dấu:
Với x < -3
Ta có: -x - 3 + 3x + 2 = x + 2
<=> 2x - 1 = x + 2
<=> x = 3 ( ko t/mãn )
Với -3 ≤ x < -2/3
Ta có: x + 3 + 3x + 2 = x + 2
<=> 4x + 5 = x + 2
<=> 3x = -3
<=> x = -1 ( t/mãn )
Với -2/3 ≤ x
Ta có: x + 3 - 3x - 2 = x + 2
<=> -2x + 1 = x + 2
<=> -3x = 1
<=> x = -1/3 ( t/mãn )
Vậy....
d, \(\left||x-1|-5\right|=x+5\)
Đk: x + 5 ≥ 0 => x ≥ -5
\(\Rightarrow\orbr{\begin{cases}\left|x-1\right|-5=x+5\\\left|x-1\right|-5=-x-5\end{cases}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=x+25\\\left|x-1\right|=-x\left(Loai\right)\end{cases}}}\)
Giải \(\left|x-1\right|=x+25\)
\(\Rightarrow\orbr{\begin{cases}x-1=-x-25\\x-1=x+25\end{cases}\Rightarrow\orbr{\begin{cases}2x=-24\\0x=26\left(Loai\right)\end{cases}\Rightarrow x}=-12}\)( ko t/mãn )
Vậy x \(\in\varnothing\)
