taco
|x+1|\(\ge\)0
|x+2|\(\ge\)0
|x+3|\(\ge\)0
\(\Rightarrow\)4x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
tu do ta co x+1+x+2+x+3=4x
3x+6=4x
4x-3x=6
x=6
tìm x biết: lx+1l+lx+2l+lx+3l=4x
Tìm x biết
lx+1l + lx+2l + lx+3l = 4x
giup minh nha
1,Tìm x,biết
a,lx+2l+lx+3l+l2-6xl=72-x
b,lx-2l-lx+3l=lx+7l
Giúp mik vs nha
tìm x biết lx-2l+lx-3l+lx-4l =2
tìm x biết rằng: lxl + lx+1l+lx+2l+lx+3l=6x
Tìm x:
a)lx-2l + lx-3l + l2x-8l =9
b)lx-1l + 3lx-3l - 2lx-2l = 4
Tìm x biết :
a; lx+2l+lx+4l+...+lx+10l=6x
b; lx+1l+lx+2l+...+lx+10l=13x-26
c; l1-xl+l2-xl+l3-xl=x-5
d;lx+1/2l+lx-1/3l+lx+1/4l=10-5x
bn nào pk lm giúp mk vs ạ
lx+1/2l+lx+1/3l+lx+1/4l=4x
lx+1l+lx+2l+lx+3l=4x
