Bài 17:
Ta có:
\(P=\dfrac{15}{x^2-2x+4}\)
\(P=\dfrac{15}{x^2-2x+1+3}\)
\(P=\dfrac{15}{\left(x^2-2x+1\right)+3}\)
\(P=\dfrac{15}{\left(x-1\right)^2+3}\)
Mà: \(\left(x-1\right)^2\ge0\forall x\) nên:
\(P=\dfrac{15}{\left(x-1\right)^2+3}\le5\forall x\)
Dấu "=" xảy ra:
\(\dfrac{15}{\left(x-1\right)^2+3}=5\)
\(\Leftrightarrow\left(x-1\right)^2+3=3\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy: \(P_{max}=5\) khi \(x=1\)
Bài 18:
Ta có:
\(Q=\dfrac{18}{4x-x^2-7}\)
\(Q=\dfrac{18}{-x^2+4x-7}\)
\(Q=\dfrac{18}{-\left(x^2-4x+4\right)-3}\)
\(Q=\dfrac{18}{-\left(x-2\right)^2-3}\)
Mà: \(-\left(x-2\right)^2\le0\forall x\) nên:
\(Q=\dfrac{18}{-\left(x-2\right)^2-3}\ge-6\forall x\)
Dấu "=" xảy ra:
\(\dfrac{18}{-\left(x-2\right)^2-3}=-6\)
\(\Leftrightarrow-\left(x-2\right)^2-3=-3\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(Q_{min}=-6\) khi \(x=2\)
17: x^2-2x+4=x^2-2x+1+3=(x-1)^2+3>=3
=>\(P=\dfrac{15}{x^2-2x+4}=\dfrac{15}{\left(x-1\right)^2+3}< =\dfrac{15}{3}=5\)
Dấu = xảy ra khi x=1
18: -x^2+4x-7
=-(x^2-4x+7)
=-(x^2-4x+4+3)
=-(x-2)^2-3<=-3
=>\(Q=\dfrac{18}{4x-x^2-7}=\dfrac{18}{-\left(x-2\right)^2-3}>=\dfrac{18}{-3}=-6\)
Dấu = xảy ra khi x=2
