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x<-1=>x+1<=0
|x+1|=-x-1
x<-1=>2-x>1>0
|2-x|=2-x
N=2-x+3(x+1)=2-x+3x+3=2x+5
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Tính từng phép tính trong ngoặc ta được :
\(A= \frac{3}{4}. \frac{8}{9} . ....\frac{899}{900}\)
\(A=\frac{1.3}{2.2} .\frac{2.4}{3.3}.... \frac{29.31}{30.30}\)
Gộp các thừa số với sau được
\(A= \frac{(1.2.3.4....29)(3.4.5.6...31)}{(2.3.4...30)(2.3.4..30)}\)
\(A= \frac{31}{30.2} = \frac{31}{60}\)
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<=>x2+2x+1+3(x2+5x-5x-25)-(4x2-4x+1)
<=>x2+2x+1+3x2+15x-15x-75-4x2+4x-1
<=>6x=75
k mình nha bạn
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Xét \(x< \frac{1}{4}\Rightarrow\hept{\begin{cases}4x-1< 0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=1-4x\\\left|x-3\right|=3-x\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(1-4x\right)=2x+5\)
Xét \(\frac{1}{4}\le x< 3\Rightarrow\hept{\begin{cases}4x-1\ge0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=3-x\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(4x-1\right)=-6x+7\)
\(x\ge3\Rightarrow\hept{\begin{cases}4x-1>0\\x-3\ge0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=x-3\end{cases}}\)
Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)=-2x-5\)
TH1: Nếu \(x< \frac{1}{4}\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=-\left(4x-1\right)\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)+\left(4x-1\right)\)
\(=-2x+6+4x-1=2x+5\)
TH2: Nếu \(\frac{1}{4}\le x\le3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=4x-1\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)-\left(4x-1\right)\)
\(=-2x+6-4x+1=-6x+7\)
TH3: Nếu \(x>3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=x-3\\\left|4x-1\right|=4x-1\end{cases}}\)
\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)\)
\(=2x-6-4x+1=-2x-5\)
Lời giải:
Nếu $x\geq 1$ thì:
$|x-1|-|x+2|=(x-1)-(x+2)=-3$
Nếu $-2\leq x< 1$ thì:
$|x-1|-|x+2|=1-x-(x+2)=-1-2x$
Nếu $x<-2$ thì:
$|x-1|-|x+2|=1-x+x+2=3$