x2-7x+8=x2-7x-7-1=(x2-1)-(7x+7)=(x-1)(x+1)-7(x+1)=(x-8)(x+1)
3:
a)
*Đa thức \(A=x\left(2x-3\right)\)
Ta có: \(A=x\left(2x-3\right)\)
\(=2x^2-3x\)
\(=2\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)\)
\(=2\left[\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}\right)-\frac{9}{4}\right]\)
\(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\)
\(=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow2\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge\frac{-9}{2}\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow\)\(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy: GTNN của đa thức \(A=x\left(2x-3\right)\) là \(\frac{-9}{2}\) khi \(x=\frac{3}{2}\)
*Đa thức \(B=x\left(x-3\right)\)
Ta có: \(B=x\left(x-3\right)\)
\(=x^2-3x\)
\(=x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\)
\(=\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}\right)-\frac{9}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge\frac{-9}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy: GTNN của đa thức \(B=x\left(x-3\right)\) là \(\frac{-9}{4}\) khi \(x=\frac{3}{2}\)
b)
* Đa thức \(C=-x^2+13x+2012\)
Ta có: \(C=-x^2+13x+2012\)
\(=-\left(x^2-13x-2012\right)\)
\(=-\left(x^2-2\cdot x\cdot\frac{13}{2}+\frac{169}{4}-\frac{8217}{4}\right)\)
\(=-\left[\left(x^2-2\cdot x\cdot\frac{13}{2}+\frac{169}{4}\right)-\frac{8217}{4}\right]\)
\(=-\left[\left(x-\frac{13}{2}\right)^2-\frac{8217}{4}\right]\)
\(=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\)
Ta có: \(\left(x-\frac{13}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow-\left(x-\frac{13}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-\frac{13}{2}\right)^2=0\Leftrightarrow\left(x-\frac{13}{2}\right)^2=0\Leftrightarrow x=\frac{13}{2}\)
Vậy: GTLN của đa thức \(C=-x^2+13x+2012\) là \(\frac{8217}{4}\) khi \(x=\frac{13}{2}\)
*Đa thức \(D=-x^2+2x-3\)
Ta có: \(D=-x^2+2x-3\)
\(=-\left(x^2-2x+3\right)\)
\(=-\left(x^2-2x+1+2\right)\)
\(=-\left[\left(x^2-2x+1\right)+2\right]\)
\(=-\left[\left(x-1\right)^2+2\right]\)
\(=-\left(x-1\right)^2-2\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-1\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-1\right)^2-2\le-2\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: GTLN của đa thức \(D=-x^2+2x-3\) là -2 khi x=1
