\(K=-x^2+10x-30\)
\(=-x^2+10x-25-5\)
\(=-\left(x^2-10x+25\right)-5\)
\(=-\left(x-5\right)^2-5\)
Ta thấy: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-5\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-5\right)^2-5\le-5\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy \(K_{max}=-5\) khi \(x=5.\)
\(---\)
\(Q=-25y^2-20y-2\)
\(=-25y^2-20y-4+2\)
\(=-\left(25y^2+20y+4\right)+2\)
\(=-\left(5y+2\right)^2+2\)
Ta thấy: \(\left(5y+2\right)^2\ge0\forall y\)
\(\Rightarrow-\left(5y+2\right)^2\le0\forall y\)
\(\Rightarrow-\left(5y+2\right)^2+2\le2\forall y\)
Dấu \("="\) xảy ra \(\Leftrightarrow5y+2=0\Leftrightarrow y=-\dfrac{2}{5}\)
Vậy \(Q_{max}=2\) khi \(y=-\dfrac{2}{5}.\)
#\(Toru\)
\(K=-x^2+10x-30\\ =-\left(x^2-10x+25\right)-5\\ =-\left(x^2-2.x.5+5^2\right)-5\\ =-\left(x-5\right)^2-5\le-5\forall x\in R\\ Vậy:max_K=-5.khi.x=5\)
\(Q=-25y^2-20y-2\\ =-\left(25y^2+20y+4\right)+2\\ =-\left[\left(5y\right)^2+2.5y.2+2^2\right]+2\\ =-\left(5y+2\right)^2+2\le2\forall y\in R\\ Vậy:max_Q=2.khi.y=-\dfrac{2}{5}\)
