Tìm cực trị của:
\(B=\dfrac{x^4+1}{\left(x^2+1\right)^2}\)
\(D=\left|11^m-5^m\right|\)
Nhắc nhẹ tí! Giải hộ đuy. :3
\(B=\dfrac{\left(x^2+1\right)^2-2x^2}{\left(x^2+1\right)^2}=1-\dfrac{2x^2}{\left(x^2+1\right)^2}\)
Do \(\dfrac{2x^2}{\left(x^2+1\right)^2}\ge0\Rightarrow B\le1\Rightarrow B_{max}=1\) khi \(\dfrac{2x^2}{\left(x^2+1\right)^2}=0\Leftrightarrow x=0\)
\(B=\dfrac{1}{2}+\dfrac{\dfrac{1}{2}x^4-x^2+\dfrac{1}{2}}{\left(x^2+1\right)^2}=\dfrac{1}{2}+\dfrac{\left(x^2-1\right)^2}{2\left(x^2+1\right)^2}\)
Do \(\dfrac{\left(x^2-1\right)^2}{\left(x^2+1\right)^2}\ge0\Rightarrow B\ge\dfrac{1}{2}\)
\(\Rightarrow B_{min}=\dfrac{1}{2}\) khi \(\dfrac{\left(x^2-1\right)^2}{\left(x^2+1\right)^2}=0\Rightarrow x=\pm1\)
\(D=\left|11^m-5^m\right|\ge0\) (theo tính chất trị tuyệt đối)
\(\Rightarrow D_{min}=0\) khi \(11^m-5^m=0\Leftrightarrow11^m=5^m\Leftrightarrow m=0\)
Không tồn tại giá trị \(D_{max}\)
\(2B=\dfrac{2x^4+2}{x^4+2x^2+1}\)
\(2B-1=\dfrac{x^4-2x^2+1}{x^4+2x^2+1}=\dfrac{\left(x^2-1\right)^2}{\left(x^2+1\right)^2}\ge0\)
=> \(B\ge\dfrac{1}{2}\)
Dấu "=" xảy ra <=> \(x^2-1=0\) <=> \(x=\pm1\)
\(D=\left|11^m-5^m\right|\ge0\)
Dấu "=" xảy ra <=> m = 0
ối lạy! lớp 8 có khái niệm "cực trị" rồi sao
