Bài 4: Giá trị tuyệt đối của một số hữu tỉ. Cộng, trừ, nhân, chia số thập phân

\(C=-\left|x+\dfrac{5}{3}\right|\\ Ta.có:\left|x+\dfrac{5}{3}\right|\ge0\forall x\in R\\ \left|x+\dfrac{5}{3}\right|_{\left(Min\right)}=0\left(khi:x=-\dfrac{5}{3}\right)\\ Vậy:C_{max}=-\left|x+\dfrac{5}{3}\right|=0\left(khi:x=-\dfrac{5}{3}\right)\)

\(D=2-\left|3-x\right|\\ Vì:\left|3-x\right|\ge0\forall x\in R\\ \Rightarrow D=2-\left|3-x\right|\le2\forall x\in R\\ Vậy:D_{max}=2\left(khi:x=3\right)\)

Ta có: \(\left|A\right|\ge0\forall x\in R\) với A là đa thức chứa x

Từ đó: 

\(A=2\cdot\left|x-2\right|+3\ge2\cdot0+3=3\\ \Rightarrow A_{min}=3\Leftrightarrow x=2\)

\(B=\left|x-3\right|+3\ge3\)

\(\Rightarrow B_{min}=3\Leftrightarrow x=3\)

\(A=2\left|x-2\right|+3\)

TH1: \(\left|x-2\right|=x-2\) với \(x-2\ge0\Leftrightarrow x\ge2\)

Nên : \(A=2\left(x-2\right)+3=2x-4+3=2x-1\)

TH2: \(\left|x-2\right|=-\left(x-2\right)\) với \(x-2< 0\Leftrightarrow x< 2\)

Nên: \(A=-2\left(x-2\right)+3=-2x+4+3=-2x+7\)

a: =>x-2=0 và y+3=0

=>x=2 và y=-3

b: =>|x-2|=|x+3|

=>x-2=x+3 hoặc x+3=2-x

=>2x=-1

=>x=-1/2

c: TH1: x<-5/4

Pt sẽ là -x-5/4+3/4-x=1

=>-2x-1/2=1

=>-2x=3/2

=>x=-3/4(loại)

TH2: -5/4<=x<3/4

Pt sẽ là x+5/4+3/4-x=1

=>8/4=1(loại)

TH3: x>=3/4

Pt sẽ là x-3/4+x+5/4=1

=>2x+1/2=1

=>2x=1/2

=>x=1/4(loại)

Ta có: \(\left|x+5\right|-\left|1-2x\right|=x\)

TH1: 

\(\left|x+5\right|=x+5\) với \(x+5\ge0\Leftrightarrow x\ge-5\)

\(\left|1-2x\right|=1-2x\) với \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

Pt trở thành: 

\(\left(x+5\right)-\left(1-2x\right)=x\)  (ĐK: \(-5\le x\le\dfrac{1}{2}\))

\(\Leftrightarrow x+5-1+2x=x\)

\(\Leftrightarrow3x+4=x\)

\(\Leftrightarrow3x-x=-4\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\left(tm\right)\)

TH2: 

\(\left|x+5\right|=-\left(x+5\right)\) với \(x+5< 0\Leftrightarrow x< -5\)

\(\left|1-2x\right|=-\left(1-2x\right)\) với \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

Pt trở thành: 

\(-\left(x+5\right)--\left(1-2x\right)=x\) (ĐK: \(\dfrac{1}{2}< x< -5\)) 

\(\Leftrightarrow-x-5+1-2x=x\)

\(\Leftrightarrow-3x-4=x\)

\(\Leftrightarrow-4x=4\)

\(\Leftrightarrow x=-1\left(ktm\right)\)

TH3: 

\(\left|x+5\right|=x+5\) với \(x+5\ge0\Leftrightarrow x\ge-5\)

\(\left|1-2x\right|=-\left(1-2x\right)\)với \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

Pt trở thành:

\(x+5--\left(1-2x\right)=x\) (ĐK: \(x>\dfrac{1}{2}\)

\(\Leftrightarrow x+5+1-2x=x\)

\(\Leftrightarrow-x+6=x\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\left(tm\right)\)

TH4: 

\(\left|x+5\right|=-\left(x+5\right)\) với \(x+5< 0\Leftrightarrow x< -5\)

\(\left|1-2x\right|=1-2x\) với \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

Pt trở thành: 

\(-\left(x+5\right)-\left(1-2x\right)=x\) (ĐK: \(x< -5\)

\(\Leftrightarrow-x-5-1+2x=x\)

\(\Leftrightarrow x-6=x\)

\(\Leftrightarrow-6=0\left(ktm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

=>|x+5|-|2x-1|=x

TH1: x<-5

Pt sẽ là -x-5-(1-2x)=x

=>-x-5-1+2x=x

=>x-6=x(loại)

TH2: -5<=x<1/2

Pt sẽ là x+5-1+2x=x

=>3x+4-x=0

=>x=-2(nhận)

TH3: x>=1/2

=>x+5-2x+1=x

=>-x+6-x=0

=>x=3(nhận)

a) TH1: Với \(x< 0\) thì \(\left|2x+3\right|=-\left(2x+3\right)=-2x-3\)

TH2: Với \(x\ge0\) thì \(\left|2x+3\right|=2x+3\)

b) TH1: Với \(x< 0\) thì \(\left|4x-2\right|=-\left(4x-2\right)=-4x+2\)

TH2: Với \(x\ge0\) thì \(\left|4x-2\right|=4x-2\)

c) TH1: Với \(x< 0\) thì \(\left|3x-5\right|=-\left(3x-5\right)=-3x+5\)

TH2: Với \(x\ge0\) thì \(\left|3x-5\right|=3x-5\)

a: TH1: x>=-3/2

=>A=2x+3

TH2: x<-3/2

=>A=-2x-3

b: TH1: x>=1/2

=>A=4x-2

TH2: x<1/2

=>A=-4x+2

c: TH1: x>=5/3

=>B=5x-3

TH2: x<5/3

=>B=-5x+3

\(TH1:\left|x+1\right|=x+1\\ \Leftrightarrow2x-\left(x+1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x-x-1=-\dfrac{1}{2}\\ \Leftrightarrow x-1=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{2}+1\\ \Leftrightarrow x=\dfrac{1}{2}\\ TH2:\left|x+1\right|=-x-1\\ \Leftrightarrow2x-\left(-x-1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x+x+1=-\dfrac{1}{2}\\ \Leftrightarrow3x=-\dfrac{1}{2}-1\\ \Leftrightarrow3x=-\dfrac{3}{2}\\ \Leftrightarrow x=\left(-\dfrac{3}{2}\right):3=-\dfrac{1}{2}\)

Thay lần lượt \(x=\dfrac{1}{2};x=-\dfrac{1}{2}\) vào pt

\(\Rightarrow x=\dfrac{1}{2}\left(thoaman\right)\)

Vậy \(x=\dfrac{1}{2}\)

\(2x-\left|x+1\right|=-\dfrac{1}{2}\)

\(\left|x+1\right|=2x+\dfrac{1}{2}\)

\(\left[{}\begin{matrix}x+1=2x+\dfrac{1}{2}\\x+1=-2x-\dfrac{1}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\).

2x  - | x + 1| = \(-\dfrac{1}{2}\) ⇒ 2x + \(\dfrac{1}{2}\) = | x + 1| đk 2x + \(\dfrac{1}{2}\) ≥ 0 ⇒ x ≥ - \(\dfrac{1}{4}\)

Với x ≥ \(-\dfrac{1}{4}\) ta có : 2x + \(\dfrac{1}{2}\) = x + 1 ⇒  2x - x = 1 - \(\dfrac{1}{2}\) ⇒ x = \(\dfrac{1}{2}\) (thỏa mãn)

vậy x  = \(\dfrac{1}{2}\)

\(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\\ =>\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{3}+\dfrac{1}{6}\\\dfrac{1}{2}x=-\dfrac{1}{3}+\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\div\dfrac{1}{2}\\x=-\dfrac{1}{6}\div\dfrac{1}{2}\end{matrix}\right.\)

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