a) Ta có: \(x\left(3-x\right)+1=3x-x^2+1\)
\(=-x^2+3x+1=-\left(x^2-3x-1\right)\)
\(=-\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{9}{4}-1\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy: Giá trị lớn nhất của biểu thức \(x\left(3-x\right)+1\) là \(\frac{13}{4}\) khi \(x=\frac{3}{2}\)
b) Ta có: \(x^2-6x+11\)
\(=x^2-6x+9+2=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
\(\Rightarrow\frac{-1}{\left(x-3\right)^2+2}\ge\frac{-1}{2}\forall x\)
hay \(A=\frac{-1}{x^2-6x+11}\ge-\frac{1}{2}\forall x\)
Dấu '=' xảy ra khi \(\left(x-3\right)^2+2=2\)
hay \(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: Giá trị nhỏ nhất của biểu thức \(A=\frac{-1}{x^2-6x+11}\) là \(-\frac{1}{2}\) khi x=3
a, Ta có : \(x\left(3-x\right)+1\)
= \(3x-x^2+1\)
= \(-\left(x^2-3x-1\right)\)
= \(-\left(x^2-2.x.\frac{3}{2}+2,25-3,25\right)\)
= \(-\left(\left(x-1,5\right)^2-3,25\right)\)
= \(3,25-\left(x-1,5\right)^2\)
Ta thấy : \(\left(x-1,5\right)^2\ge0\forall x\)
=> \(-\left(x-1,5\right)^2\le0\)
=> \(3,25-\left(x-1,5\right)^2\le3,25\)
- Dấu " = " xảy ra khi \(x-1,5=0\)
=> \(x=1,5\)
Vậy Max = 3,25 khi x = 1,5 .
