Bài 5: Phương trình chứa dấu giá trị tuyệt đối

\(a^4+1\ge a\left(a^2+1\right)\Leftrightarrow a^3\left(a-1\right)-\left(a-1\right)\ge0\) \(\Leftrightarrow\left(a^3-1\right)\left(a-1\right)\ge0\) 

\(\Leftrightarrow\left(a-1\right)^2\left(a^2+a+1\right)\ge0\) (Luôn đúng)

\(a^4+1\ge a\left(a^2+1\right)\\ \Leftrightarrow a^4+1\ge a^3+a\\ \Leftrightarrow a^4+1-a-a^3\ge0\Leftrightarrow\left(a^4-a\right)-\left(a^3-1\right)\ge0\\ \Leftrightarrow a\left(a-1\right)\left(a^2+a+1\right)-\left(a-1\right)\left(a^2+a+1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a^3+a^2+a-a^2-a-1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a^3-1\right)\ge0\\ \Leftrightarrow\left(a-1\right)\left(a-1\right)\left(a^2+a+1\right)\ge0\)

Nhận xét:

(a - 1)(a - 1) = (a - 1)² \(\ge\) 0 ∀ x

\(a^2+a+1\\ =a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy \(\left(a-1\right)^2\left(a^2+a+1\right)\) luôn dương 

=> ... (đpcm)

f: 2|x+1|-3|x-3|=3

TH1: x<-1

=>-2x-2-3(3-x)=3

=>-2x-2-9+3x=3

=>x-11=3

=>x=14(loại)

TH2: -1<=x<3

Pt sẽ là 2(x+1)-3(3-x)=3

=>2x+2-9+3x=3

=>5x-7=3

=>x=2(nhận)

TH3: x>=3

=>2(x+1)-3(x-3)=3

=>2x+2-3x+9=3

=>-x+11=3

=>x=8(nhận)

a: =>|5x-2|=|2x-3|

=>5x-2=2x-3 hoặc 5x-2=-2x+3

=>3x=-1 hoặc 7x=5

=>x=5/7 hoặc x=-1/3

b: =>|5x-2|-|2x+2|=3x+5

TH1 x<-1

PT sẽ là 2-5x+2x+2=3x+5

=>-3x+4=3x+5

=>-6x=1

=>x=-1/6(loại)

TH2: -1<=x<2/5

Pt sẽ là 2-5x-2x-2=3x+5

=>-7x=3x+5

=>-4x=5

=>x=-5/4(loại)

Th3: x>=2/5

PT sẽ là 5x-2-2x-2=3x+5

=>3x-4=3x+5

=>0x=9(loại)

\(\Leftrightarrow2x-3=7hay2x-3=-7\)

\(\Leftrightarrow2x=10hay2x=-4\)

\(\Leftrightarrow x=5hayx=-2\)

=> 2x - 3 = 7          hoặc 2x -3 = -7  

     2x = 3 + 7                   2x = 3 +(-7)

     2x = 10                        2x = -4

       x = 10 : 2                    x = -4 : 2

      x=5                                  x=-2

a: TH1: x<1

Pt sẽ là 1-x+3-x=7

=>4-2x=7

=>2x=-3

=>x=-3/2(nhận)

TH2: 1<=x<3

Pt sẽ là x-1+3-x=7

=>2=7(loại)

Th3: x>=3

PT sẽ là x-1+x-3=7

=>2x-4=7

=>x=11/2(nhận)

b: x^2+4=4|x|(2)

TH1: x>=0

(2) =>x^2+4=4x

=>x^2-4x+4=0

=>x=2(nhận)

Th2: x<0

(2) =>x^2+4=-4x

=>x^2+4x+4=0

=>x=-2(nhận)

c: =>x-7>=0

=>x>=7

a.\(\left|x-3\right|=4x+1\)

 \(ĐK:4x+1\ge0\Leftrightarrow4x\ge-1\Leftrightarrow x\ge\dfrac{-1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4x+1\\x-3=-4x-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4x=1+3\\x+4x=-1+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x=4\\5x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4}{3}\left(ktm\right)\\x=\dfrac{2}{5}\left(tm\right)\end{matrix}\right.\)

Vay S \(=\left\{\dfrac{2}{5}\right\}\)

b. \(\left|x-2\right|+2x=10\\ \Leftrightarrow\left|x-2\right|=10-2x\)

ĐK : \(10-2x\ge0\Leftrightarrow-2x\ge-10\Leftrightarrow x\le5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10-2x\\x-2=2x-10\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2x=10+2\\x-2x=-10+2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=12\\-x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=8\left(ktm\right)\end{matrix}\right.\)

Vay S \(=\left\{4\right\}\)

/2-x/ = 7 
<=> 2-x = +- 7 
xét 2 th 
th1 : 2-x = 7                            2-x = -7 
           x = -5                               x= 9         

| 2 - x | = 7

@ TH1: 2 - x = 7

     <=> x = 9

@ TH2: 2 - x = -7

    <=> x = -5

Vậy x = -5 hoặc x = 9

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

\(ĐK:x\le\dfrac{3}{2}\)

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

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

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

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

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

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

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

\(\Leftrightarrow x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=3\\-2x-5=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=-2\\-2x=8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

Giúp mình với

|2x – 5| = 3

TH1: |2x – 5| = 2x – 5 khi 2x – 5 ≥ 0 ó 2x ≥ 5 ó x ≥ 5/2 

Khi đó |2x – 5| = 3

=> 2x – 5 = 3 ó 2x = 8 ó x = 4 (TM)

TH2: |2x – 5| = - (2x – 5) khi 2x – 5 < 0 ó 2x < 5 ó x < 5/2 

Khi đó |2x – 5| = 3

ð - (2x – 5) = 3 ó 2x = 2 ó x = 1 (TM)

Vậy phương trình có nghiệm là x = 4; x = 1