Tìm x, biết 3.|x-1|=0

Bài 1:

a) 3.|x-1|=0

x-1=0

x=0+1=1

b) 5.|x-1|+3=8

=> 5.|x-1|=8-3=5

x-1=5:5=1

x=1+1=2

c) |x+1|:2+1=3

|x+1|=(3-1).2=4

=>\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-4+1=-3\end{matrix}\right.\)

d) |5-x|.3+1=4

|5-x|.3 =4-1=3

|5-x| =3:3=1

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

e) |7+x| : 2 = 6

|7+x| = 6 . 2 = 12

\(\Rightarrow\left[{}\begin{matrix}7+x=12\\7+x=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12-7=5\\x=-12-7=-19\end{matrix}\right.\)

f) |x-5| + 3 = 4

|x-5| =4 - 3 = 1

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

g) |7-x| +3 = 3

|7-x| = 3 - 3 = 0

=> 7-x =0

x=0+7=7

h) |8+x| : 8 = 2

|8+x| =2.8=16

=> \(\left[{}\begin{matrix}8+x=16\\8+x=-16\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8-16=-8\\x=8-\left(-16\right)=8+16=24\end{matrix}\right.\)

i) |10-x| . 3 = 3

|10-x| = 3 : 3 = 1

=> \(\left[{}\begin{matrix}10-x=1\\10-x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-1=9\\x=10+1=11\end{matrix}\right.\)

k) \(\left|2-x\right|:2=4\\ \left|2-x\right|=4.2=8\\ \Rightarrow\left[{}\begin{matrix}2-x=8\\2-x=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-8=-6\\x=2+8=10\end{matrix}\right.\)

Bài 2: Hình bạn tự vẽ nha

a) Vì xOy < xOz (30^o < 60^o) nên tia Oy nằm giữa hai tia Ox và Oz.

b) Do tiaOy nằm giữa hai tia còn lại nên xOz = xOy + yOz

=> yOz = xOz - xOy = 60^o - 30^o = 30^o

^{Chúc bạn học tốt}