\(B=x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(B_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
\(C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(C_{max}=7\) khi \(x=2\)
\(D=-\dfrac{1}{2}\left(4x^2-4x+1\right)-\dfrac{9}{2}=-\dfrac{1}{2}\left(2x-1\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(D_{max}=-\dfrac{9}{2}\) khi \(x=\dfrac{1}{2}\)
\(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
giá trị nhỏ nhất là 3/4
b)
\(C=4x-x^2+3\)
\(=-x^2+4x+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\)
vậy giá trị lớn nhất là 7
c)
\(D=2x-2x^2-5\)
\(=-2x^2+2x-5\)
\(=-\left(2x^2-2x+5\right)\)
\(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
vậy giá trị nhỏ nhất là -9/4
