Question

tìm GTLN của A=\(\frac{1}{x^2-1}+\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}\)

Answer 1

\(A=\frac{1}{\left(x-1\right)\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}\)

\(2A=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}=\frac{1}{x-1}-\frac{1}{x+5}\)

\(2A=\frac{x+5-x+1}{\left(x-1\right)\left(x+5\right)}=\frac{6}{x^2+4x-5}\Leftrightarrow A=\frac{3}{\left(x+2\right)^2-9}\le\frac{3}{-9}=-3\)

Max A = -3 khi x =-2 (TM)