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Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
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|\(\left|x+\frac{1}{3}\right|+\left|x+\frac{4}{3}\right|=\frac{9}{2}.x\) (1)
+)Xét \(x< -\frac{4}{3}\);(1) trở thành: \(-2x-\frac{5}{3}=\frac{9}{2}x\Leftrightarrow\frac{13}{2}x=-\frac{5}{3}\Leftrightarrow x=-\frac{10}{39}\)(loại)
+)Xét \(-\frac{4}{3}\le x< -\frac{1}{3}\);(1) trở thành: \(\frac{9}{2}x=1\Leftrightarrow x=\frac{2}{9}\) (loại)
+)Xét \(x>-\frac{1}{3}\);(1) trở thành: \(2x+\frac{5}{3}=\frac{9}{2}x\Leftrightarrow\frac{5}{2}x=\frac{5}{3}\Leftrightarrow x=\frac{2}{3}\) (chọn)
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`4x(x-5)-(x-1) (4x-3)-5=0`
`=> 4x*x - 4x*5 - ( x*4x-3*x-1*4x+ 1*3) -5=0`
`=> 4x^2 - 20x-(4x^2 -3x-4x+3)-5=0`
`=> 4x^2 - 20x-4x^2+3x+4x-3-5=0`
`=>-13x-8=0`
`=> -13x=8`
`=> x=-8/13`
Vậy `x=-8/13`
`4x(x-5)-(x-1)(4x-3)-5 = 0`
`=> 4x^2 - 20x - (4x^2 -3x-4x+3)= 5`
`=> 4x^2 - 20x - 4x^2 + 3x + 4x -3 = 5`
`=> (4x^2 - 4x^2) - (20x - 3x - 4x) = 8`
`=> -13x = 8`
`=> x = -8/13`
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\(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)
\(2x-3-x+5=x+2-x+1\)
\(x+2=3\)
\(x=3-2\)
\(x=1\)