e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

1 

e) E >= 2021 

dấu = xảy ra khi x=1/2

g) G = |x-1|+ |2-x| >= |x-1+2-x|=1

Dấu = xảy ra khi (x-1)(2-x)>=0 <=> 1<=x<=2

h) H = |x-1|+|x-2| + |x-3| 

Ta có : |x-1| + |x-3| = |x-1| + |3-x| >= |x-1+3-x| = 2

|x-2| >=0

=> H>=2

Dấu = xảy ra khi (x-1)(3-x) >=0 ; x-2=0

<=> x=2

k) K = |x-1| + |2x-1| 

2K = |2x-2| + |2x-1| + |2x-1|

Ta có : |2x-2| + |2x-1|  = |2x-2| + |1-2x| >= |2x-2+1-2x|=1

|2x-1| >=0 

Dấu = xảy ra (2x-2)(1-2x) >=0; 2x-1=0

<=> x=1/2

e)Vì \(\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|\ge0\forall x\\ \Rightarrow2\left|x-\dfrac{1}{2}\right|+2012\ge2012\forall x\)

Dấu "=" xảy ra khi x=\(\dfrac{1}{2}\)

Vậy...

b)G=|x-1|+ |2-x|\(\)

áp dụng bđt |a+b|+ |c+d|\(\ge\left|a+b+c+d\right|\forall x\)

\(\Rightarrow\)ta có |x-1|+ |2-x|\(\ge\) \(\left|x-1+2-x\right|\forall x\)

\(\Leftrightarrow\text{|x-1|+ |2-x| }\ge1\forall x\)

Dấu "=" xảy ra khi 1\(\le x\le2\) \(\forall x\)

Vậy...

h)H= |x-1|+|x-2| + |x-3| 

Ta có |x-1| + |x-3|         

=|x-1| + |3-x| ( trong giá trị tuyệt đối đổi dấu không cần đặt dấu trừ ở ngoài)       

 =>|x-1| + |3-x|\(\ge\left|x-1+3-x\right|\forall x\)          

<=>|x-1| + |3-x|\(\ge2\forall x\) (1)

Mà |x-2|\(\ge0\forall x\) (2)

Từ (1) và (2)=> ta có |x-1|+|x-2| + |x-3| \(\ge2\forall x\)

Dấu "=" xảy ra khi x-2=0

<=>x=2

Vậy...

k) K = |x-1| + |2x-1| 

2K = |2x-2| + |2x-1| + |2x-1|

Mà : |2x-2| + |2x-1| 

=|2x-2| + |1-2x|\(\ge\text{|2x-2+1-2x|}\) \(\forall x\)

Lại có |2x-1| \(\ge\)0 \(\forall x\)

Dấu "=" xảy ra 2x-1=0

<=>x=\(\dfrac{1}{2}\)

Vậy....

a.

\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

\(x=-\frac{1}{2}\)

TH2:

\(x-\frac{3}{4}=0\)

\(x=\frac{3}{4}\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)

b.

\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)

TH1:

\(\frac{1}{2}x-3=0\)

\(\frac{1}{2}x=3\)

\(x=3\div\frac{1}{2}\)

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

\(\frac{2}{3}x=-\frac{1}{2}\)

\(x=-\frac{1}{2}\div\frac{2}{3}\)

\(x=-\frac{1}{2}\times\frac{3}{2}\)

\(x=-\frac{3}{4}\)

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)

\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)

\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)

\(x=5\)

\(\Leftrightarrow5+3x^2+9x< 3x^2+6x-x-2\)

\(\Leftrightarrow9x-6x+x< 3x^2-3x^2-5-2\)

\(\Leftrightarrow2x< -7\)

\(\Leftrightarrow x< \frac{-7}{2}\)

10 - { [ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 } = 5

[ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 = 10 - 5 = 5

( x : 3 + 17 ) : 10 + 3 : 2⁴ = 5 x 10

( x : 3 + 17 ) : 10 + 48 = 50

( x : 3 + 17 ) : 10 = 50 - 48

( x : 3 + 17 ) : 10 = 2

x : 3 + 17 = 2 x 10

x : 3 + 17 = 20

x : 3 = 20 - 17 = 3

x = 3 x 3 = 9

a) [(2x+14) : 4 - 3] : 2 = 1

(2x+14) : 4 - 3 = 1/2

(2x+14) : 4  = 1/2 + 3

(2x+14) : 4  = 7/2

2x+14 = 7/2 . 1/4

2x = 7/8 - 1/4

2x = 5/8

x= 5/8.1/2

x= 5/16

Còn phần b nữa ak

Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)

\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)

Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)

      \(\left|x-\frac{3}{7}\right|\ge0\forall x\)

Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)