Bài 1:
\(x^2-5x-6=0\)
\(\Leftrightarrow x^2+x-6x-6=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(6x+6\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)
Vậy x=-1; x=6
Bài 2:
a) Ta có: \(x+y=10\Leftrightarrow y=10-x\) (1)
Từ (1) thay vào \(P=xy\) ta được:
\(P=x\left(10-x\right)\)
\(\Leftrightarrow P=10x-x^2\)
\(\Leftrightarrow P=-x^2+10x-5^2+5^2\)
\(\Leftrightarrow P=-\left(x^2-10x+5^2\right)+5^2\)
\(\Leftrightarrow P=-\left(x-5\right)^2+25\)
Vậy GTLN của P=25 khi \(x-5=0\Leftrightarrow x=5\)
b) \(P=x^2-5x\)
\(\Leftrightarrow P=x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\)
\(\Leftrightarrow P=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\)
Vậy GTNN của \(P=\dfrac{-25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)