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\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)
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\(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right).x+\frac{4}{325}+1+\left(\frac{5}{324}+1\right)+\frac{349}{5}-4=0\)
\(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right).x=-329\)
=>x=-329
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a, Xét : x-4 = 0 => x= 4
2x+1 = 0 => x= \(\frac{1}{2}\)
x+3 = 0 => x = -3
x + 9 = 0 => x = -9
Khi đó ta có bảng xét dấu :
x | -9 | -3 | \(\frac{1}{2}\) | 4 |
x-4 | -13 | -7 | \(\frac{-7}{2}\) | 0 |
2x+1 | -17 | -5 | 2 | 9 |
x+3 | -6 | 0 | \(\frac{7}{2}\) | 7 |
x+9 | 0 | 6 | \(\frac{19}{2}\) | 13 |
=> có 5 trường hợp:
TH1 : \(x\le-9\)
TH2 : \(-9\le x< -3\)
TH3 : \(-3\le x< \frac{1}{2}\)
TH4 : \(\frac{1}{2}\le x< 4\)
Do đó :
TH1 : \(x\le-9\)
Ta có : /x-4/ = -(x-4) = 4 - x
/2x+1/ = -(2x+1) = -2x -1
/x+3/ = -(x + 3 ) = -x - 3
/x-9/ = -(x-9) = -x + 9 Thay vào đề bài ta có:
3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5
=> 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5
=>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5
=> -13 - 7x = 5
7x = -13 - 5
7x = -18
x = \(\frac{-18}{7}\)( Ko TM)
Tương tự với 4 trường hợp còn lại.
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1440:[(-4)-(2.x-5)]=24:22
1440:[(-4)-(2.x-5)]=24-2
1440:[(-4)-(2.x-5)]=22
1440:[(-4)-(2.x-5)]=4
-4-(2.x-5)=1440:4
-4-(2.x-5)=360
2.x-5=-4-360
2.x-5=-364
2.x=-364+5
2.x=-359
x=-359:2
x=-179,5
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\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)
Th1:
\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)
\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)
( 1 )
Th2:
\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)
\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)
( 2 )
Từ ( 1 ) và ( 2 ), ta có:
\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)
\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
Th1:
\(2-x>0\Leftrightarrow x>2\)
\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)
( Loại )
Th2:
\(2-x< 0\Leftrightarrow x< 2\)
\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)
=> \(\dfrac{4}{5}< x< 2\)