Question

[HanSoo >>>^^^.^^^<<<]

Tìm giá trị lớn nhất :

A = -|x - 2| + 2019

B = -2x² + 5x + 3

C = -x² - y² + 2x + 8y + 2028

Ai nhanh và đúng mk tick chooo ~!~!~!~!~!

Answer 1

[lê duy mạnh]

a<=2019 dấu = xảy ra khi x=2

Answer 2

[Lê Tài Bảo Châu]

a) \(A=-|x-2|\le0;\forall x\)

\(\Rightarrow-|x-2|+2019\le0+2019;\forall x\)

Hay \(A\le2019;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow|x-2|=0\)

                        \(\Leftrightarrow x=2\)

Vậy \(A_{max}=2019\Leftrightarrow x=2\)

b)  \(B=-2x^2+5x+3\)

  \(=-2\left(x^2-\frac{5}{2}x-\frac{3}{2}\right)\)

\(=-2\left(x^2-2.x.\frac{5}{4}+\frac{25}{16}-\frac{25}{16}-\frac{3}{2}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\)

Vì \(-2\left(x-\frac{5}{4}\right)^2\le0;\forall x\)

\(\Rightarrow-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\le0+\frac{49}{8};\forall x\)

Hay \(B\le\frac{49}{8};\forall x\)

Dấu "="xảy ra \(\Leftrightarrow\left(x-\frac{5}{4}\right)^2=0\)

                       \(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(B_{max}=\frac{49}{8}\Leftrightarrow x=\frac{5}{4}\)

c) \(-x^2-y^2+2x+8y+2028\)

\(=-\left(x^2+y^2-2x-8y-2028\right)\)

\(=-\left[\left(x^2-2x+1\right)+\left(y^2-8y+16\right)-2045\right]\)

\(=-\left(x-1\right)^2-\left(y-4\right)^2+2045\)

Vì \(\hept{\begin{cases}-\left(x-1\right)^2\le0;\forall x,y\\-\left(y-4\right)^2\le0;\forall x,y\end{cases}}\)

\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2\le0;\forall x,y\)

\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2+2045\le0+2045;\forall x,y\)

Hay \(C\le2045;\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}-\left(x-1\right)^2=0\\-\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}}\)

Vậy \(C_{max}=2045\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}\)