Bài 2:
$a)5(x+4)-3(x-2)=x\\\Leftrightarrow 5x+20-3x+6=x\\\Leftrightarrow 2x+26=x\\\Leftrightarrow 2x-x=-26\\\Leftrightarrow x=-26$
Vậy $x=-26$
$b) \left | x+4 \right |=2^{0}+1^{2003}\\\Leftrightarrow \left | x+4 \right |=1+1\\\Leftrightarrow \left | x+4 \right |=2\\\Leftrightarrow \left[ \begin{array}{l}x+4=2\\x+4=-2\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-2\\x=-6\end{array} \right.$
Vậy $x\in \left \{ -6;-2 \right \}$
c) Ta có: $\left | x+8 \right |+\left | x+8 \right |+\left | x+1963 \right |\geq \left | x+8+x+8+x+1963 \right |=\left | 3x+1979 \right |$
Mà $\left | x+8 \right |+\left | x+8 \right |+\left | x+1963 \right |=4x$

$\Rightarrow \left | 3x+1979 \right |=4x\\\Leftrightarrow \left | 3x+1979 \right |-4x=0\\\Leftrightarrow \left[ \begin{array}{l}3x+1979-4x=0,3x+1979\geq 0\\-(3x+1979)-4x=0,3x+1979<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1979,x\geq \dfrac{1979}{3}\\x=-\dfrac{1979}{7},x<\dfrac{1979}{3}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1979\\x\in \varnothing \end{array} \right.$
Vậy $x=1979$

Đáp án:

 a) x=-26

b)

c)

Giải thích các bước giải:

 a) 5.(x+4) - 3(x-2) = x

⇒ 5x+20 - 3x + 6 = x

⇒ 5x - 3x - x = -20 - 6

⇒ x=-26

b) |x+4| = $2^{0}$ + $1^{2003}$ 

⇒ |x+4| = 1 + 1

⇒ |x+4| = 2

⇒ $\left \{ {{x+4=2} \atop {x-4=-2}} \right.$ 

⇒$\left \{ {{x=-2} \atop {x=-6}} \right.$ 

c) khó quá mik ko biết làm