Question

Cho biểu thức: \(M=\left(\dfrac{x^2-1}{x^4-x^2+1}-\dfrac{1}{x^2+1}\right).\left(x^4+\dfrac{1-x^4}{1+x^2}\right)\)

a) Rút gọn.

b) Tìm GTNN của M.

Answer 1

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

\(\Rightarrow M\ge-2\)