(2x-1)^3 =-8

=>(2x-1)³=(-2)³

=>2x-1=-2

=>2x=-2+1

=>2x=-1

=>x=-1/2

<=>±x-3=±-3-x

=>x-3=3-x

để 2 vế = nhau thì kq fai =0

=>x=3

thử lại x-3=0=3-3

x=3

k cho mink nhé

ta có:|x-3|=|3-x|

ta có: |x-3|=-|x-3|

|x-3|+|x-3|=0

2|x-3|=0

x-3=0

x=3

|3x - 1| = 3²

=> |3x - 1| = 9

=> 3x - 1 = 9 hoặc 3x - 1 = -9

=> 3x = 10 hoặc 3x = -8

=> x = 10/3 hoặc x = -8/3

Mà x < 0

=> x = -8/3

a, \(\left|4x-8\right|\le8\)

\(\Leftrightarrow\left(\left|4x-8\right|\right)^2\le64\)

\(\Leftrightarrow16x^2-64x+64\le64\)

\(\Leftrightarrow16x^2-64x\le0\)

\(\Leftrightarrow16x\left(x-4\right)\le0\)

\(\Leftrightarrow0\le x\le4\)

b, \(\left|x-5\right|\le4\)

\(\Leftrightarrow\left(\left|x-5\right|\right)^2\le16\)

\(\Leftrightarrow x^2-10x+25\le16\)

\(\Leftrightarrow x^2-10x+9\le0\)

\(\Leftrightarrow1\le x\le9\)

\(\Rightarrow x\in\left\{1;2;3;4;5;6;7;8;9\right\}\)

c, \(\left|2x+1\right|< 3x\)

TH1: \(x\ge-\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow2x+1< 3x\)

\(\Leftrightarrow x>1\)

\(\Rightarrow\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

TH2: \(x< -\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow-2x-1< 3x\)

\(\Leftrightarrow x>-\dfrac{1}{5}\left(l\right)\)

Vậy \(\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

d, \(\left|x+1\right|+\left|x\right|< 3\)

\(\Leftrightarrow x+1+x+2\left|x^2+x\right|< 9\)

\(\Leftrightarrow\left|x^2+x\right|< 4-x\)

Xét hai trường hợp để phá dấu giá trị tuyệt đối

e, Tương tự câu d

|x|=1<=>x=1 hoặc x=-1

+)x=1=>2x⁴+3x²+5=2.1⁴+3.1²+5=10

+)x=-1=>2x⁴+3x²+5=2.(-1)⁴+3.(-1)²+5=10

Vậy A=10

\(A=\left|x-3\right|+\left|y+3\right|+2016\)

\(\left|x-3\right|\ge0\)

\(\left|y+3\right|\ge0\)

\(\Rightarrow\left|x-3\right|+\left|y+3\right|+2016\ge2016\)

Dấu ''='' xảy ra khi \(x-3=y+3=0\)

\(x=3;y=-3\)

\(MinA=2016\Leftrightarrow x=3;y=-3\)

\(\left(x-10\right)+\left(2x-6\right)=8\)

\(x-10+2x-6=8\)

\(3x=8+10+6\)

\(3x=24\)

\(x=\frac{24}{3}\)

x = 8