\(2\left(x+\dfrac{1}{x}\right)^2=2\left(x^2+\dfrac{1}{x^2}+2\right)=2\left(x^2+\dfrac{1}{x^2}\right)+4\)
Chắc em nhìn thấy hằng đẳng thức trong căn rồi, có cần làm tiếp không?
Ủa đâu phải, ngoài dấu căn mà:
\(B=\left(-x^2+x-1\right):\sqrt{\left(x^2+\dfrac{1}{x^2}\right)^2+2\left(x^2+\dfrac{1}{x^2}\right)+1}\)
\(=-\left(x^2-x+1\right):\sqrt{\left(x^2+\dfrac{1}{x^2}+1\right)^2}\)
\(=-\left(x^2-x+1\right):\left(x^2+\dfrac{1}{x^2}+1\right)\)
\(=-\dfrac{\left(x^2-x+1\right).x^2}{x^4+x^2+1}=-\dfrac{\left(x^2-x+1\right)x^2}{x^4+2x^2+1-x^2}\)
\(=-\dfrac{\left(x^2-x+1\right)x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=-\dfrac{x^2}{x^2+x+1}\)
\(B=\dfrac{-3x^2}{3\left(x^2+x+1\right)}=\dfrac{-4\left(x^2+x+1\right)+x^2+4x+4}{3\left(x^2+x+1\right)}=-\dfrac{4}{3}+\dfrac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\ge-\dfrac{4}{3}\)
