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a) \(ĐKXĐ:m\ne0,m\ne\pm1\)
Ta có : \(P=\left(\frac{1+m}{m\left(m-1\right)}\right):\frac{m+1}{\left(m-1\right)^2}\)
\(=\frac{1+m}{m\left(m-1\right)}\cdot\frac{\left(m-1\right)^2}{m+1}\)
\(=\frac{m-1}{m}\)
Vây \(P=\frac{m-1}{m}\) thỏa mãn ĐKXĐ.
b) Khi \(m=\frac{1}{2}\) ( thỏa mãn ĐKXĐ ) thì \(P=\frac{\frac{1}{2}-1}{\frac{1}{2}}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}.2=1\)
Vậy : \(P=1\) khi \(m=\frac{1}{2}\)
Ta có : \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{a^3b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{a^2b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{ab}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{1}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{1}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{1}\right)\)
=> \(P=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2+2a\right)-6a}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a+b\right)^2}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}+\frac{6}{\left(a+b\right)^4}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}=\frac{2a^2+4ab+2b^2}{\left(a+b\right)^2}-\frac{a^2+b^2}{\left(a+b\right)^2}\)
=> \(P=2-\frac{a^2+b^2}{\left(a+b\right)^2}=1+\frac{-2}{\left(a+b\right)^2}\)
\(VT=\frac{t}{\left(m-1\right)\left(m^2+m+1\right)}-\frac{m}{\left(t-1\right)\left(t^2+t+1\right)}=\frac{t}{-t\left(m^2+m+1\right)}-\frac{m}{-m\left(t^2+t+1\right)}\)\(=\frac{-1}{m^2+m+1}+\frac{1}{t^2+t+1}=\frac{-t^2-t-1+m^2+m+1}{\left(m^2+m+1\right)\left(t^2+t+1\right)}\)
\(=\frac{\left(m-t\right)\left(m+t\right)+m-t}{m^2t^2+mt\left(m+t\right)+m^2+t^2+mt+\left(m+t\right)+1}\)
\(=\frac{2\left(m-t\right)}{m^2t^2+\left(m^2+t^2+2mt\right)+2}=\frac{2\left(m-t\right)}{m^2t^2+\left(m+t\right)^2+2}=\frac{2\left(m-t\right)}{m^2t^2+3}=VP\)