Question

Tìm GTLN của các biểu thức sau:

a. \(4x-x^2-1\)

b.\(3x-3x^2-7\)

c.\(x+6y-x^2-y^2-10\)

d.\(-\left(3x+7\right)^2+2\left(3x+7-17\right)\)

Answer 1

a: \(-x^2+4x-1\)

\(=-\left(x^2-4x+1\right)\)

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5< =5\)

Dấu '=' xảy ra khi x=2

b: \(-3x^2+3x-7\)

\(=-3\left(x^2-x+\dfrac{7}{3}\right)\)

\(=-3\left(x^2-x+\dfrac{1}{4}+\dfrac{25}{12}\right)\)

\(=-3\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\le-\dfrac{25}{4}\)

Dấu '=' xảy ra khi x=1/2

d: \(=-\left(3x+7\right)^2+2\left(3x+7\right)-34\)

\(=-\left[\left(3x+7\right)^2-2\left(3x+7\right)+34\right]\)

\(=-\left[\left(3x+7\right)^2-2\left(3x+7\right)+1+33\right]\)

\(=-\left(3x+7-1\right)^2-33\)

\(=-\left(3x+6\right)^2-33\le-33\)

Dấu '=' xảy ra khi x=-2