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![](https://rs.olm.vn/images/avt/0.png?1311)
Tiếp
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(\frac{x^2+x+1}{2x+1}\right)=\left(\frac{x^2+x+1}{x^2-1}\right)=1+\frac{x+2}{x^2-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
\(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x+1}{x\left(x-1\right)}\)
\(=\frac{x^2}{x-1}\)
Vì \(P=-\frac{1}{2}\)
\(\Leftrightarrow\frac{x^2}{x-1}=-\frac{1}{2}\)ĐKXĐ:\(x\ne1\)
\(\Rightarrow2x^2=-x+1\)
\(\Rightarrow x^2+x^2+x-1=0\)
\(\Rightarrow x\left(x+1\right)+\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Rightarrow\)\(x=0\) \(\Rightarrow\)\(x=0\)(TM)
\(x+1=0\) \(x=-1\)(TM)
\(x-1=0\) \(x=1\)(KTM)
Vậy để \(P=-\frac{1}{2}\)thì x=0 hoặc x=-1