a) \(M=\left(\dfrac{x^2-1}{x^4-x^2+1}-\dfrac{1}{x^2+1}\right).\left(x^4+\dfrac{1-x^4}{1+x^2}\right)\)
\(M=\left(\dfrac{(x^2-1)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{(x^4-x^2+1)\left(x^2+1\right)}\right).\left(\dfrac{x^6+x^4+1-x^4}{1+x^2}\right)\)
\(M=\left(\dfrac{x^4-1-\left(x^4-x^2+1\right)}{x^6+1}\right).\left(\dfrac{x^6+1}{1+x^2}\right)\)
\(M=\left(\dfrac{x^2-2}{x^6+1}\right).\left(\dfrac{x^6+1}{1+x^2}\right)\)
\(M=\dfrac{x^2-2}{x^2+1}\)
b) Ta có:
\(x^2\ge0\left(\forall x\right)\)
\(x^2-2\ge-2\)
\(\dfrac{x^2-2}{x^2+1}\ge-2\)
do mẫu \(x^2+1\) lớn hơn 0 nên chia ko cần đổi dấu
\(\Rightarrow M\ge-2\)
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