a) Để đẳng thức xảy ra thì: 101x\(\ge\)0=>x\(\ge\)0

Suy ra: \(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)

<=>\(100x+\frac{1+2+....+100}{101}=101x\)

<=>x=\(\frac{\frac{\left(1+100\right).100}{2}}{101}=50\)

phan b lap bang xet dau

x -5 0,5

x+5 0 /

1-2x / 0

voi x<-5 ta co pt: -x-5-1+2x=x

-6=0(loai)

voi -5=<x=<0,5 :x+5-1+2x=x

2x=-2(nhan)

voi x>0,5: x+5+1-2x=x

x=3(nhan)

b)

TH1:x<-5 suy ra |x+5|=-x-5; |1-2x|=1-2x

ta có:(-x-5)-(1-2x)=x

-x-5-1+2x=x

-2x=6

x=-3(loại)

TH2:-5</x<1/2 suy ra |x+5|=x+5; |1-2x|=1-2x

ta có:(x+5)-(1-2x)=x

x+5-1+2x=x

2x=4

x=2(loại)

TH3:x>/1/2 suy ra |x+5|=x+5; |1-2x|=2x-1

ta có: (x+5)-(2x-1)=x

x+5-2x+1-x=0

-2x=-4

x=2(chọn)

vậy x=2 thỏa mãn yêu cầu đề

no bít hahaha....?

a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)

c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)

2|2x - 3| = 1/2

=> |2x - 3| = 1/4

=> 2x - 3 = 1/4 hoặc 2x - 3 = -1/4

đến đây dễ bn tự tính được

um mk chiu

a) |x+1|+|x+2+|x+3|=4x

<=> x+1+x+2+x+3=4x

<=> 3x+6=4x

<=> 6=4x-3x

<=> x=6

b) |x+1|+|x+2|+|x+3|+|x+4|=5x

<=> x+1+x+2+x+3+x+4=5x

<=> 4x+10=5x

<=> 10=5x-4x

<=> x=10