b/\(\Leftrightarrow\frac{m\left(x+m\right)}{x^2-m^2}-\frac{3m^2-4m+3}{x^2-m^2}=\frac{x-m}{x^2-m^2}\)
\(\Leftrightarrow mx+m^2-3m^2+4m-3=x-m\)
\(\Leftrightarrow-2m^2+mx+5m-x-3=0\)
\(\Leftrightarrow\left(-2m^2+2m+3m-3\right)+x\left(m-1\right)=0\)
\(\Leftrightarrow-2m\left(m-1\right)+3\left(m-1\right)+x\left(m-1\right)=0\)
\(\Leftrightarrow\left(m-1\right)\left(x-2m+3\right)=0\Rightarrow\left[{}\begin{matrix}m=1\left(1\right)\\x=2m-3\left(2\right)\end{matrix}\right.\)
(ĐKXĐ x khác +-m)
-Với (1) PT đúng với mọi x
-Với (2), PT TM khi \(x=2m-3\ne+-m\Leftrightarrow\left\{{}\begin{matrix}m-3\ne0\\3m-3\ne0\end{matrix}\right.\)
Vậy (2) là nghiệm khi m khác (3,1)
a/Chuyển vế ta xét biểu thức
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}=\frac{a^2+b^2+2}{\left(1+a^2\right)\left(1+b^2\right)}-\frac{2}{1+ab}=\frac{\left(a^2+b^2+1\right)\left(1+ab\right)-2\left(1+a^2\right)\left(1+b^2\right)}{\left(1+ab\right)\left(1+a^2\right)\left(1+a^2\right)}=\frac{a^2+b^2+2+\left(a^2+b^2\right)ab+2ab-2-2a^2-2b^2-2a^2b^2}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}=\frac{\left(a^2+b^2\right)ab-\left(a^2+b^2\right)+2ab\left(1-ab\right)}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}=\frac{\left(a^2+b^2\right)\left(ab-1\right)-2ab\left(ab-1\right)}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}=\frac{\left(ab-1\right)\left(a-b\right)^2}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}\left(1\right)\)
Có a,b >= 1 nên (1) không âm, suy ra BĐT luôn đúng với mọi a,b>=1
