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ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x^2+x+1\ne0\end{cases}}\)
a/ \(R=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right]\)
\(=1:\left[\frac{x^2+2+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x}{x^2+x+1}\right)\)
\(=\frac{x^2+x+1}{x}\)
b/ Ta có: \(R=\frac{x^2+x+1}{x}=3+\frac{\left(x-1\right)^2}{x}>3\)
Vậy R > 3
a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)
B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)
=\(\frac{3-x^2}{\left(x-1\right)^2}\)
b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)
TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)
c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2
I don't now
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ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
a: \(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x\left(x^2+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x^2+1\right)}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
Để R=0 thì \(x^2+1=0\)(vô lý)
b: Ta có: |x|=1
=>x=1(loại) hoặc x=-1(loại)
a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)
\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{x+1}{x}\right)+\frac{1}{\left(x+1\right)^2}.\left(\frac{x^2+1}{x^2}\right)\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{2}{\left(x+1\right)^2x}+\frac{x^2+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{x^2+2x+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{3}\)
\(P=\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}:\frac{x-1}{3}\)
\(P=\frac{3}{x^2\left(x-1\right)}\)
b)Bài này liên quan đến dấu lớn nên mk ko làm đc
