Question

Tìm giá trị lớn nhất của

M=\(-4x^2+x+1\)

N=\(4-x^2+3x\)

Answer 1

M=-4x²+x+1

=-4(x²-\(\frac{1}{4}\)x)+1

=-4(x²-2.\(\frac{1}{8}\)x+\(\frac{1}{64}\))+1-(-4.\(\frac{1}{64}\))

=-4(x-\(\frac{1}{8}\))²+\(\frac{17}{16}\le\frac{17}{16}\)

Dấu = xảy ra : -4(x-\(\frac{1}{8}\))²=0

\(\Leftrightarrow x=\frac{1}{8}\)

N=4-x²+3x

=-(x²-3x)+4

=-(x²-2.\(\frac{3}{2}\)x+\(\frac{9}{4}\))+4-(-\(\frac{9}{4}\))

=-(x-\(\frac{3}{2}\))²+\(\frac{25}{4}\le\frac{25}{4}\)

Dấu = xảy ra khi : -(x-\(\frac{3}{2}\))²=0

\(\Leftrightarrow x=\frac{3}{2}\)

Answer 2

\(N=4-x^2+3x=-\left(x^2-3x+4\right)=-\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{7}{4}\right)\)

\(=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le\frac{-7}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)