Question

Tìm giá trị lớn nhất

1>   y = 4x - x²

2>   y = 2 - 3x - x²

3 >   y = 4 - 6x - x²

4>   y = 3 - 6x -2x²

5> y = \(\frac{x-1}{x^2}\)

Answer 1

Bài 1: 

\(y=-\left(x^2-4x\right)=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Bài 2: 

\(y=-\left(x^2+3x-2\right)\)

\(=-\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{17}{4}\right)\)

\(=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\forall x\)

Dấu '=' xảy ra khi x=-3/2

Bài 3: 

\(y=-\left(x^2+6x-4\right)\)

\(=-\left(x^2+6x+9-13\right)\)

\(=-\left(x+3\right)^2+13\le13\forall x\)

Dấu '=' xảy ra khi x=-3