a.
\(3-\left(17-x\right)=289-\left(36+289\right)\)
\(3-17+x=289-36-289\)
\(x=\left(289-289\right)+\left(-36-3+17\right)\)
\(x=-22\)
b.
\(x-73=\left(-35\right)+\left|-55\right|\)
\(x-73=-35+55\)
\(x=-35+55+73\)
\(x=93\)
c.
\(\left|x\right|-3=2\)
\(\left|x\right|=3+2\)
\(\left|x\right|=5\)
\(x=\pm5\)
Vậy x = 5 hoặc x = -5
d.
\(\left|x+1\right|=5\)
\(x+1=\pm5\)
TH1:
\(x+1=5\)
\(x=5-1\)
\(x=4\)
TH2:
\(x+1=-5\)
\(x=-5-1\)
\(x=-6\)
Vậy x = 4 hoặc x = -6
a, \(3-17+x=289-36-289\)
\(\Rightarrow x-14=-36\)
\(\Rightarrow x=-22\)
\(b,x-73=\left(-35\right)+\left|-55\right|\)
\(\Rightarrow x-73=\left(-35\right)+55\)
\(\Rightarrow x-73=20\)
\(\Rightarrow x=93\)
\(c,\left|x\right|-3=2\)
\(\Rightarrow\left|x\right|=5\)
\(\Rightarrow x=\pm5\)
\(d,\left|x+1\right|=5\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+1=5\\x+1=-5\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=4\\x=-6\end{array}\right.\)