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

Ta có :

\(N=\left|x-2014\right|+\left|2015-x\right|\ge\left|x-2014+2015-x\right|=1\)

Dấu "=" xảy ra khi :

\(\Leftrightarrow\left(x-2014\right)\left(2015-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2014\ge0\\2015-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2014\le0\\2015-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2014\\2015\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2014\\2015\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2014\le x\le2015\\x\in\varnothing\end{matrix}\right.\)

Vậy \(N_{Min}=1\Leftrightarrow2014\le x\le2015\)

N= | x-2014 | +|2015 -x| ≥| x-2014 + 2015 -x | = | 1| = 1

dấu "=" khi x-2014 = 2015 - x

<=> x = 2014,5

vậy gtnn N = 1 khi x = 2014,5

\(\left|x+\dfrac{1}{2}\right|-\dfrac{2}{5}=0\)

\(\Rightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{2}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\pm\dfrac{2}{5}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

Vậy..............

\(\left|x+\dfrac{1}{2}\right|-\dfrac{2}{5}=0\)

\(\left|x+\dfrac{1}{2}\right|=\dfrac{2}{5}\)

=> \(x+\dfrac{1}{2}=\dfrac{2}{5}\) Hoặc \(x+\dfrac{1}{2}=\dfrac{-2}{5}\)

Với : \(x+\dfrac{1}{2}=\dfrac{2}{5}\)

=> \(x=\dfrac{2}{5}-\dfrac{1}{2}\)

\(x=\dfrac{-1}{10}\)

Với : \(x+\dfrac{1}{2}=\dfrac{-2}{5}\)

\(x=\dfrac{-2}{5}-\dfrac{1}{2}\)

\(x=\dfrac{-9}{10}\)

Vì |x+\(\dfrac{1}{2}\)|-\(\dfrac{2}{5}\)=0

Nên x hoặc -x

X=\(\dfrac{2}{5}-\dfrac{1}{2}\)=\(\dfrac{4}{9}\)

=>x=\(\dfrac{4}{9}\)hoặc\(\dfrac{-4}{9}\)

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

\(\Rightarrow\dfrac{1}{2}x+2=3\)

\(\Rightarrow\dfrac{1}{2}x=1\)

\(\Rightarrow x=2\)

Vậy ...........

Từ \(\left|\dfrac{1}{2}x+2\right|=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x+2=3\\\dfrac{1}{2}x+2=-3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=3-2\\\dfrac{1}{2}x=-3-2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1:\dfrac{1}{2}\\x=-5:\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

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

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

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=1\\\dfrac{1}{2}x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

vậy \(x\in\left\{2,-10\right\}\)

a)\(\left(\left|2-3x\right|+\left|x+1\right|\right)^2=9\)

\(\Leftrightarrow\left(2-3x\right)^2+\left(x+1\right)^2+2\left|\left(2-3x\right)\left(x+1\right)\right|=9\)

\(\Leftrightarrow10x^2-10x+5+2\left|-3x^2-x+2\right|=9\)

\(\Leftrightarrow\left|-3x^2-x+2\right|=\dfrac{4+10x-10x^2}{2}=2+5x-5x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x^2-x+2=2+5x-5x^2\\-3x^2-x+2=-2-5x+5x^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-6x=0\\8x^2-4x^2-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

vậy \(S=\left\{-\dfrac{1}{2};0;1;3\right\}\)

b) Trường hợp 1: x+6>0 và x-5>0 suy ra x>5

\(\Leftrightarrow x+6+x-5=3\)

\(\Leftrightarrow x=1\)

Trường hợp 2 x+6<0 và x-5>0 ( vô lí)

Trường hợp 3: x+6>0 và x-5<0 \(\Leftrightarrow-6< 0< 5\)

\(\Leftrightarrow x+6-\left(x-5\right)=3\)

\(11=3\) ( vô lí)

Trường hợp 4 x+6<0 và x-5<0 suy ra x<-6

\(\Leftrightarrow-\left(x+6\right)-\left(x-5\right)=3\)

\(\Leftrightarrow x=-2\)

vậy \(S=\left\{-2;1\right\}\)

a) Vì \(\left|2-3x\right|\) và \(\left|x+1\right|\) luông\(\ge\) 0 với mọi x, y \(\in\) Z.
Mà \(\left|2-3x\right|\)+\(\left|x+1\right|\) =0
\(\Rightarrow\left\{{}\begin{matrix}2-3x=0\\x+1=0\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}3x=2-0\\x=0-1\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}3x=2\\x=-1\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=2:3\\x=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-1\end{matrix}\right.\)
Vậy x\(\in\)\(\left\{\dfrac{2}{3};-1\right\}\)
Tick cho mình nha!!!

a)/4.x/-12.5=7.3

/4.x/-60=21

/4.x/=21+60

/4.x/=81

4.x=81 hoặc 4.x=-81

x=81:4 hoặc x=-81:4

x=20,25 hoặc x=-20,25

vậy ....

b)/x/+x=1/3

x+x=1/3

2x=1/3

x=1/3:2

x=1/6

=>/x/=1/6

=>x=+-1/6

c)/x/-x=3/4

TH1 x>=0 TH2 x<0

=>/x/-x khác 3/4 =>/x/-x=3/4

=>x thuộc rỗng =>/x/+x=3/4

=>2x=3/4=>x=3/8

d)/x-2/=x

=>x-2=+-x

=>x-2=x(vô lí) x-2=-x

ta thử có x=1 t/m yêu cầu

=>x =1

e)/x+2/=x

x+2=+-x

x+2=x(không t/m) x+2=-x

ta thử có x=-1 t/m

vậy x=-1

Ta có 2 TH :

+TH1: \(x\ge0\) => \(x-\left|x\right|=x-x=0\)

+TH2: \(x< 0\) => \(x-\left|x\right|=x-\left(-x\right)=x+x=2x\)

Vậy giá trị lớn nhất là 2x

Gọi đẳng thức trên là A

\(\left|x+1\right|\) + \(\left|2x-4\right|\)

Vì \(\left|x+1\right|\) \(\ge\) 0

\(\Rightarrow\left|x+1\right|\) + \(\left|2x-4\right|\)\(\ge\) 0 + \(\left|2x-4\right|\)

\(\Rightarrow\) A \(\ge\) \(\left|2x-4\right|\)

Dấu "=" xảy ra khi \(\left|x+1\right|\) = \(\left|2x-4\right|\)

\(\left|x+1\right|\) = 2x - 4

\(\Rightarrow\) x + 1 = 2x-4 hoặc x + 1 = -(2x -4)

2x - x = -4 - 1 x = -(2x-4)-1

x = -5 x = (2x -5).(-1)

x:(-1) = 2x-5

-x = 2x-5

2x-(-x)=5

3x =5

x = \(\dfrac{5}{3}\)

Vậy GTNN của A là \(\left|2x-4\right|\) khi x=-5

x=\(\dfrac{5}{3}\)

\(\left|x+1\right|\)

\(\left|x+1\right|\)