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\(A=\frac{x\left|x-2\right|}{x^2+8x-20}=\frac{x\left|x-2\right|}{x^2-2x+10x-20}=\frac{x\left|x-2\right|}{x\left(x-2\right)+10\left(x-2\right)}=\frac{x\left|x-2\right|}{\left(x+10\right)\left(x-2\right)}\)
Xét \(x-2\ge0\Leftrightarrow x\ge2\) ta có :
\(A=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}=\frac{x}{x+10}\)
Xét \(x-2< 0\Leftrightarrow x< 2\) ta có :
\(A=\frac{x\left(2-x\right)}{\left(x+10\right)\left(x-2\right)}=\frac{-x}{x+10}\)
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Ta có: a)(x - 5).(2x +3) - (2x -1).(x +7) - (x -1).(x+2)
= 2x2 + 3x - 10x - 15 - 2x2 - 14x + x + 7 - x2 - 2x + x + 2
= -x2 - 21x - 6
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a) + Với x < 1, ta có:
|x| + |x - 1| = -x + [-(x - 1)]
= -x - x + 1
= -2x + 1
+ Với \(x\ge1\), ta có:
|x| + |x - 1| = x + x - 1
= 2x - 1
b) + Với \(x< -\frac{2}{3}\), ta có:
|3x + 2| - (x + 1) = -(3x + 2) - (x + 1)
= -3x - 2 - x - 1
= - 4x - 3
+ Với \(x\ge-\frac{3}{2}\), ta có:
|3x + 2| - (x + 1) = 3x + 2 - x - 1
= 2x + 1
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\(A=\frac{x^2}{x^2-1}-\frac{x^2}{x^2+1}\left(\frac{x}{x+1}+\frac{1}{x^2+x}\right)\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x}{x+1}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x^2}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}.\frac{x^2+1}{x\left(x+1\right)}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x}{x+1}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
=>\(A=\frac{x^2-x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x^2-x^2+x}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x}{x^2-1}\)
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\(A=\left|x-3,5\right|+2x-7\)
Với \(x\ge3,5\)thì \(\left|x-3,5\right|=x-3,5\)
Do đó \(A=x-3,5+2x-7\)
\(\Leftrightarrow A=3x-10,5\)
Với \(x< 3,5\)thì \(\left|x-3,5\right|=3,5-x\)
Do đó \(A=3,5-x+2x-7\)
\(\Leftrightarrow A=x-3,5\)