a)\(\left(x-2\right)^2-1\)
Dễ thấy:\(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2-1\ge-1\forall x\)
Đẳng thức xảy ra khi \(x=2\)
b)\(\left(x^2-9\right)^2+\left|y-2\right|+10\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(x^2-9\right)^2\ge0\\\left|y-2\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|\ge0\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|+10\ge10\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x^2-9=0\\y-2=0\end{matrix}\right.\)\(\left\{{}\begin{matrix}x=\pm3\\y=2\end{matrix}\right.\)
c)\(\dfrac{3}{\left(x-2\right)^2+5}\)
Dễ thấy:
\(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2+5\ge5\)
\(\Rightarrow\dfrac{1}{\left(x-2\right)^2+5}\le\dfrac{1}{5}\Rightarrow\dfrac{3}{\left(x-2\right)^2+5}\le\dfrac{3}{5}\)
Đẳng thức xảy ra khi \(x-2=0\Rightarrow x=2\)
d)\(-10-\left(x-30\right)^2-\left|y-5\right|\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(x-30\right)^2\ge0\\\left|y-5\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}-\left(x-30\right)^2\le0\\-\left|y-5\right|\le0\end{matrix}\right.\)
\(\Rightarrow-\left(x-30\right)^2-\left|y-5\right|\le0\)
\(\Rightarrow10-\left(x-30\right)^2-\left|y-5\right|\le10\)
Đẳng thức xảy ra khi \(\Rightarrow\left\{{}\begin{matrix}-\left(x-30\right)^2=0\\-\left|y-5\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=30\\y=5\end{matrix}\right.\)
a) \(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2-1\ge-1\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)
\(\Rightarrow x=2\)
Vậy GTNN của bt = -1 khi x = 2.
b) \(\left(x^2-9\right)^2\ge0;\left|y-2\right|\ge0\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|\ge0\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|+10\ge10\)
Dấu "=" xảy ra khi \(\left(x^2-9\right)^2=0;\left|y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm3\\y=2\end{matrix}\right.\)
Vậy GTNN của bt = 10 khi ...
c) Vì \(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+5\ge5\)
\(\Rightarrow\dfrac{3}{\left(x-2\right)^2+5}\ge\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)
\(\Rightarrow x=2\)
Vậy GTNN của bt = \(\dfrac{3}{5}\) khi x = 2.
Trước hết thế đã.
a, Đặt \(A=\left(x-2\right)^2-1\)
Ta có: \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-1\ge-1\)
Dấu " = " khi \(\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(MIN_A=-1\) khi x = 2
b, Đặt \(B=\left(x^2-9\right)^2+\left|y-2\right|+10\)
Ta có: \(x^2\ge0\)
\(\Rightarrow x^2-9\ge-9\)
\(\Rightarrow\left(x^2-9\right)^2\ge81\)
Mà \(\left|y-2\right|\ge0\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|\ge81\)
\(\Rightarrow B=\left(x^2-9\right)^2+\left|y-2\right|+10\ge91\)
Dấu " = " khi \(\left\{{}\begin{matrix}x^2=0\\\left|y-2\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Vậy \(MIN_B=91\) khi x = 0 và y = 2
c, Đặt \(C=\dfrac{3}{\left(x-2\right)^2+5}\)
Để C lớn nhất thì \(\left(x-2\right)^2+5\) nhỏ nhất
Ta có: \(\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+5\ge5\)
\(\Rightarrow C=\dfrac{3}{\left(x-2\right)^2+5}\le\dfrac{3}{5}\)
Dấu " = " xảy ra khi \(\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(MAX_C=\dfrac{3}{5}\) khi x = 2
d, sai đề không bạn?
