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\(\left|x^2-5x+4\right|=x^2-5x+4\Leftrightarrow x^2-5x+4>0\Leftrightarrow\left(x-1\right)\left(x-4\right)\ge0\)
\(\left|x^2-5x+4\right|=5x^2-x^2-4\Leftrightarrow x^2-5x+4< 0\Leftrightarrow\left(x-1\right)\left(x-4\right)< 0\)
Với \(\left|x^2-5x+4\right|=x^2-5x+4\) thì:
\(pt\Leftrightarrow x^2-5x+4=5x-x^2-4\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow x=1;x=4\)
Với \(\left|x^2-5x+4\right|=5x-x^2-4\) thì pt luôn đúng vs \(\forall x\) thỏa mãn \(\left(x-1\right)\left(x-4\right)< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
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|5x-4|=|x+2|
=> |5x-4|=x+2
=> 5x-4=x+2 hoặc 5x-4=-(x+2)=-x-2
=> 5x-x=2+4 hoặc 5x+x=-2+4
=> 4x=6 hoặc 6x=-2
=> x=6/4 hoặc x=-2/6
=> x=1,5 hoặc x=-1/3
|5x-4|=|x+2|
=>|5x-4|=x+2
=>5x-4=x+2
=>5x-x=2+4
=>4x=6
=>x=6÷4
=>x=1,5
Vậy: x=1,5
![](https://rs.olm.vn/images/avt/0.png?1311)
/ 5x -4 / = / x+ 2 /
ta xét 2 TH
TH1 x > 0
=> x + 2 > 0
và 5x - 4 > 0
=> 5x -4 = x + 2
=. 5x - x = 4 + 2
=> 4x = 6
=> x = 3/2
TH2 tự làm nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(TH1:\left(5x-4\right)\ge0;\left(x+2\right)\ge0\)
\(\Rightarrow5x-4=x+2\)
\(\Leftrightarrow5x-x=2+4\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{6}{4}=\frac{3}{2}\)
\(TH2:\left(5x-4\right)< 0;\left(x+2\right)< 0\)
\(\Rightarrow-\left(5x-4\right)=-\left(x+2\right)\)
\(\Leftrightarrow-5x+4=-x-2\)
\(\Leftrightarrow-5x+x=-2-4\)
\(\Leftrightarrow-4x=-6\)
\(\Leftrightarrow x=\frac{-6}{-4}=\frac{3}{2}\)
\(TH3:\left(5x-4\right)\ge0;\left(x+2\right)< 0\)
\(\Rightarrow5x-4=-\left(x+2\right)\)
\(\Leftrightarrow5x-4=-x-2\)
\(\Leftrightarrow5x+x=-2+4\)
\(\Leftrightarrow6x=2\)
\(\Leftrightarrow x=\frac{2}{6}=\frac{1}{3}\)
\(TH4:\left(5x-4\right)< 0;\left(x+2\right)\ge0\)
\(\Rightarrow-\left(5x-4\right)=x+2\)
\(\Leftrightarrow-5x+4=x+2\)
\(\Leftrightarrow-5x-x=2-4\)
\(\Leftrightarrow-6x=-2\)
\(\Leftrightarrow x=\frac{-2}{-6}=\frac{1}{3}\)
Vậy \(x\in\left\{\frac{1}{3};\frac{3}{2}\right\}\)
HOK TOT
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: (x-3).(4-5x) +2 =2
=> (x-3)(4-5x) = 0
=> x-3=0 hoặc 4-5x=0
=> x=3 hoặc x=4/5
\(\left(x-3\right)\left(4-5x\right)+2=2\)
\(\Rightarrow\left(x-3\right)\left(4-5x\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
|5x + 4| = |x + 2|
+ TH1: 5x + 4 = x + 2
=> 4x + 4 = 2
=> 4x = -2
=> x = \(\frac{-1}{2}\)
+ TH2: 5x + 4 = -(x + 2)
=> 5x + 4 = -x - 2
=> 5x + 6 = -x
=> -6x = 6
=> x = -1
KL: x thuộc {\(\frac{-1}{2}\); -1}
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(\begin{cases}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\\\left|x+4\right|\ge0\end{cases}\)\(\forall x\)
Do đó, \(5x-1\ge0\Rightarrow5x\ge1\Rightarrow x\ge\frac{1}{5}\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)=5x-1\)
=> 4x + 10 = 5x - 1
=> 10 + 1 = 5x - 4x
=> x = 11
Vậy x = 11