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a, \(\left|5x-3\right|< 2\)
<=> \(2^2-\left(5x-3\right)^2>0\)
<=> \(\left(5-5x\right)\left(5x-1\right)>0\)
<=> \(\frac{1}{5}< x< 1\)
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
a ) \(\left|x\right|\ge0\)với mọi x
\(\left|2+x\right|\ge0\)với mọi x
=> \(\left|x\right|+\left|2+x\right|\ge0\)với mọi x
Mà \(\left|x\right|+\left|2+x\right|=2x\)
=> \(2x\ge0\)
=> \(x\ge0\)
=> \(\hept{\begin{cases}\left|x\right|=x\\\left|2+x\right|=2+x\end{cases}}\)
=> \(\left|x\right|+\left|2+x\right|=x+2+x=2x\)
=> \(2x+2=2x\)
=> \(2=0\)( vô lí )
Vậy \(x\in\varnothing\)
b ) \(\left|x\right|< 3\)
=> \(-3< x< 3\)
c ) \(\left|x\right|>2\)
=> \(\orbr{\begin{cases}x>2\\x< -2\end{cases}}\)
Vậy x > 2 hoặc x < - 2
d ) \(\left|2-x\right|< 3\)
=> \(-3< 2-x< 3\)
=> \(3>x-2>-3\)
=> \(5>x>-1\)
e ) \(3-\left|x+2\right|\le1\)
=> \(\left|x+2\right|\le3-1\)
=> \(\left|x+2\right|\le2\)
=> \(-2\le x+2\le2\)
=> \(-4\le x\le0\)
/x/+/2+x/=2x
Vì /x/>=0;/2+x/>=0
=> /x/+/2+x/>=0
=> 2x>=0
=> x>=0
=>/x/=x
\(\Rightarrow x+|2+x|=2x\)
\(\Rightarrow|2+x|=x\)
\(\Rightarrow\hept{\begin{cases}2+x=x\\2+x=-x\end{cases}\Rightarrow\hept{\begin{cases}2=0\left(loại\right)\\-2x=2\Rightarrow x=-1\end{cases}}}\)
\(\Rightarrow x=-1\)
b, \(|x|< 3\Rightarrow-3< x< 3\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
Lắm quá oong nội ơi
a) \(\left|5x-3\right|-3x=12\)
=> \(\left|5x-3\right|=12+3x\)
=> \(\orbr{\begin{cases}5x-3=3x+12\\3-5x=3x+12\end{cases}}\)
=> \(\orbr{\begin{cases}2x=15\\-8x=-9\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{15}{2}\\x=\frac{9}{8}\end{cases}}\)
b) \(\left|x^2-2x-4\right|+4=4x\)
=> \(\left|x^2-2x-4\right|=4x-4\)
=> \(\orbr{\begin{cases}x^2-2x-4=4x-4\\x^2-2x-4=4-4x\end{cases}}\)
=> \(\orbr{\begin{cases}x^2-6x=0\\x^2+2x-8=0\end{cases}}\)
=> \(\orbr{\begin{cases}x\left(x-6\right)=0\\x^2+4x-2x-8=0\end{cases}}\)
=> x = 0 hoặc x - 6 = 0
hoặc (x - 2)(x + 4) = 0
=> x= 0 hoặc x = 6
hoặc x - 2 = 0 hoặc x + 4 = 0
=> x = 0 hoặc x = 6
hoặc x = 2 hoặc x = -4
a)TH1:|5x-3|=5x-3
5x-3-3x=12
2x=15
x=7.5
TH2:|5x-3|=-5x+3
-5x+3-3x=12
-8x=9
x= -\(\frac{9}{8}\)
b)TH1:|x2-2x-4|=x2 -2x-4
x2-2x-4+4=4x
x2-6x=0
x(x-6)=0
=>\(\hept{\begin{cases}x=0\\x=6\end{cases}}\)
TH2:|x2-2x-4|=-(x2-2x-4)
-x2+2x+4-4=4x
-x2-2x=0
-x(x+2)=0
\(\Rightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)
\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)
\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Xét các trường hợp rồi tìm được x thôi :>
\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)
\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)
d, TT
\(\left|5x-3\right|< 2\)
\(-2< 5x-3< 2\)
\(-2< 5x-3\Leftrightarrow1< 5x\Leftrightarrow x>\frac{1}{5}\left(1\right)\)
\(5x-3< 2\Leftrightarrow5x< 5\Leftrightarrow x< 1\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{1}{5}< x< 1\)
a/ \(|5x-3|< 2\) b/ \(|3x+1>4|\) c/ \(|4-x|+2x=3\)
\(\Leftrightarrow5x-3< 2\) \(\Leftrightarrow3x+1>4\) \(\Leftrightarrow4-x+2x=3\)
\(\Leftrightarrow5x< 5\) \(\Leftrightarrow3x>3\) \(\Leftrightarrow x=-1\)
\(\Leftrightarrow x< 1\) \(\Leftrightarrow x>1\)
\(a,\left|5x-3\right|< 2\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|5x-3\right|=1\\\left|5x-3\right|=0\end{cases}}\)
\(TH1:\)\(\)
\(\left|5x-3\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=1\\5x-3=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1+3\\5x=-1+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=4\\5x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\left(\text{loại}\right)\\x=\frac{2}{5}\left(\text{loại}\right)\end{cases}}\)
\(TH2:\)
\(\left|5x-3\right|=0\)
\(\Leftrightarrow5x-3=0\)
\(\Leftrightarrow5x=0+3\)
\(\Leftrightarrow5x=3\)
\(\Leftrightarrow x=\frac{3}{5}\left(\text{loại}\right)\)
\(\text{Vậy : không tồn tại x cần tìm.}\)
\(b,\left|3x+1\right|>4\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1>4\\3x+1< -4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x>4-1\\3x< -4-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x>3\\3x< -5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x>3\div3\\x< -5\div3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x>1\\x< \frac{-5}{3}\end{cases}}\)
\(\text{Vậy : }\)\(x>1\)\(\text{hoặc}\)\(x< \frac{-5}{3}\)
\(\)