Question

[Tên Ko]

Tìm x thuộc Z.

a, 3—(17—x)=289—(36+289)

b, x—73=(—35)+|—55|

|x|—3=2

|x+1|=5

Answer 1

[Phương An]

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

Answer 2

[Nguyễn Trần An Thanh]

a, \(3-17+x=289-36-289\)

\(\Rightarrow x-14=-36\)

\(\Rightarrow x=-22\)

\(b,x-73=\left(-35\right)+\left|-55\right|\)

\(\Rightarrow x-73=\left(-35\right)+55\)

\(\Rightarrow x-73=20\)

\(\Rightarrow x=93\)

\(c,\left|x\right|-3=2\)

\(\Rightarrow\left|x\right|=5\)

\(\Rightarrow x=\pm5\)

\(d,\left|x+1\right|=5\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+1=5\\x+1=-5\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=4\\x=-6\end{array}\right.\)