a) Để A nhỏ nhất thì \(2-\left(x-1\right)^4\) lớn nhất nên ta có:
\(-\left(x-1\right)^4\le0\forall x\Rightarrow2-\left(x-1\right)^4\le2\forall x\)
\(\Rightarrow\dfrac{-1}{2-\left(x-1\right)^4}\le\dfrac{-1}{2}\forall x\)
\(\Rightarrow A\le\dfrac{-1}{2}\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^4=0\Rightarrow x=1\).
Vậy \(MIN_A=\dfrac{-1}{2}\) khi \(x=1.\)
b) Ta có \(-x^2+x-3=-\left(x^2-x+3\right)\)
\(=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{11}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{11}{4}\)
\(=\dfrac{11}{4}+\left(x-\dfrac{1}{2}\right)^2\)
........
Vậy \(MAX_B=\dfrac{11}{4}\) khi \(x=\dfrac{1}{2}.\)
a, Ta có: \(2-\left(x-1\right)^4\le2\)
\(\Rightarrow A=\dfrac{-1}{2-\left(x-1\right)^4}\le\dfrac{-1}{2}\)
Dấu " = " khi \(\left(x-1\right)^4=0\Rightarrow x=1\)
Vậy \(MAX_A=\dfrac{-1}{2}\) khi x = 1
b, \(B=-x^2+x-3\)
\(=-\left(x^2-x+3\right)\)
\(=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le\dfrac{-11}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy \(MAX_B=\dfrac{-11}{4}\) khi \(x=\dfrac{1}{2}\)
\(A=\dfrac{-1}{2-\left(x-1\right)^4}đk:2-\left(x-1\right)^4\ne0\)
\(A_{MIN}\Rightarrow2-\left(x-1\right)^4_{MAX}\)
\(\left(x-1\right)^4\ge0\)
\(2-\left(x-1\right)^4_{MAX}\Rightarrow\left(x-1\right)^4_{MIN}\)
\(\left(x-1\right)^4_{MIN}=0\)
\(\Rightarrow A_{MIN}=\dfrac{-1}{2-0}=\dfrac{-1}{2}\)