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a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)
=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]
={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]
=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]
=-x.x(x-2)(x+1)/2(2-x)x3
=(x+1)/2x
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)
..........
\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)
\(\Leftrightarrow\)\(x=-2040\)
Vậy phương trình có nghiệm là : x = -2040
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)
\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)
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