Dao Trong Luan Coordinator
03/12/2017 at 10:01-
\(A=\dfrac{3-4x}{x^2+1}=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}=\dfrac{4\left(x^2+1\right)-\left(2x+1\right)^2}{x^2+1}\)
\(=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\le4\)
The equality happens only when : \(2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)
Dao Trong Luan selected this answer. -
Nguyễn Hưng Phát 03/12/2017 at 11:26We have:A-1=\(\dfrac{3-4x}{x^2+1}-1=\dfrac{3-4x-\left(x^2+1\right)}{x^2+1}=\dfrac{3-4x-x^2-1}{x^2+1}=\dfrac{-\left(x^2+4x-2\right)}{x^2+1}=\dfrac{-\left(x+2\right)^2+6}{x^2+1}=\dfrac{-\left(x+2\right)^2}{x^2+1}+\dfrac{6}{x^2+1}\) Because:\(\dfrac{-\left(x+2\right)^2}{x^2+1}\le0\) so A-1\(\le\dfrac{6}{x^2+1}\)
\(\Rightarrow\)maxA-1=\(\dfrac{6}{x^2+1}\)\(\Leftrightarrow x=-2\Rightarrow\)maxA-1=\(\dfrac{6}{5}\)\(\Rightarrow\)maxA=\(\dfrac{11}{5}\)
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FA Liên Quân Garena 30/12/2017 at 21:47We have:A-1=3−4xx2+1−1=3−4x−(x2+1)x2+1=3−4x−x2−1x2+1=−(x2+4x−2)x2+1=−(x+2)2+6x2+1=−(x+2)2x2+1+6x2+1 Because:−(x+2)2x2+1≤0 so A-1≤6x2+1
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maxA-1=6x2+1⇔x=−2⇒maxA-1=65⇒maxA=115