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a)
\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)
\(A\ge\dfrac{3}{2}\) khi x =4
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
Ta có : \(P=2x^2-8x+1=2\left(x^2-4x\right)+1=2\left(x^2-4x+4-4\right)+1=2\left(x-2\right)^2-7\)
Vì \(2\left(x-2\right)^2\ge0\forall x\)
Nên : \(P=2\left(x-2\right)^2-7\ge-7\forall x\in R\)
Vậy \(P_{min}=-7\) khi x = 2
\(b,Q=-5x^2-4x+1\)
\(=-5\left(x^2+\dfrac{4}{5}x+\dfrac{4}{25}\right)+\dfrac{9}{5}\)
\(=-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\)
Với mọi giá trị của x ta có:
\(-5\left(x+\dfrac{2}{5}\right)^2\le0\)
\(\Rightarrow-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)
Vậy MaxQ = \(\dfrac{9}{5}\)
Để Q = \(\dfrac{9}{5}\) thì \(x+\dfrac{2}{5}=0\Rightarrow x=-\dfrac{2}{5}\)
\(c,K=x\left(x-3\right)\left(x-4\right)\left(x-7\right)\)
\(=x\left(x-7\right)\left(x-3\right)\left(x-4\right)\)
\(=\left(x^2-7x\right)\left(x^2-7x+12\right)\)
Đặt \(x^2-7x+6=t\) , ta có:
\(K=\left(t-6\right)\left(t+6\right)\)
\(=t^2-36\)
\(=\left(x^2-7x+6\right)^2-36\)
Với mọi giá trị của x ta có:
\(\left(x^2-7x+6\right)^2\ge0\Rightarrow\left(x^2-7x+6\right)^2-36\ge-36\)
Vậy Min K = -36
Để K = - 36 thì \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-x-6x+6=0\)
\(\Leftrightarrow x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a)\(P=2x^2-8x+1\)
=\(2\left(x^2-4x+4\right)-7\)
=\(2\left(x-2\right)^2-7\)
Với mọi x thì \(2\left(x-2\right)^2>=0\)
=>\(2\left(x-2\right)^2-7>=-7\)
Hay \(P>=-7\) với mọi x
Để \(P=-7\) thì
\(\left(x-2\right)^2=0\)
=>\(x-2=0\)
=>\(x=2\)
Vậy...
Các câu sau tương tự
\(E=\dfrac{3}{-x^2+2x-4}=\dfrac{3}{-\left(x^2-2x+1\right)-3}=\dfrac{3}{-\left(x-1\right)^2-3}\)Vì \(-\left(x-1\right)^2-3\le-3\Rightarrow\dfrac{3}{-\left(x-1\right)^2-3}\ge\dfrac{3}{-3}=-1\)Vậy \(Min_E=-1\) khi \(x-1=0\Rightarrow x=1\)