Question

Tìm giá trị lớn nhất 

A= 15-8x-x2

B=4x-x2+2

C= x2 - y2+4x+4y+2

Answer 1

\(A=15-8x-x^2=-\left(x+4\right)^2+31\)

Vì \(\left(x+4\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+4\right)^2+31\le31\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy maxA = 31 <=> x = - 4

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

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-2\right)^2+6\le6\)

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

Vậy maxB = 6 <=> x = 2

Answer 2

a) \(A=15-8x-x^2=-\left(x^2+8x+16\right)-1\)

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

Dấu "=" xảy ra khi: \(-\left(x+4\right)=0\Rightarrow x=-4\)

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

\(=-\left(x-2\right)^2+6\le6\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)

c) Trang nghĩ nên sửa đề nhé:

\(C=-x^2-y^2+4x+4y+2\)

\(C=-\left(x^2-4x+4\right)-\left(y^2-4y+4\right)+10\)

\(C=-\left(x-2\right)^2-\left(y-2\right)^2+10\le10\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\left(x-2\right)^2=0\\-\left(y-2\right)^2=0\end{cases}}\Rightarrow x=y=2\)

Answer 3

A = 15 - 8x - x²

= -( x² + 8x + 16 ) + 31

= -( x + 4 )² + 31 

-( x + 4 )² ≤ 0 ∀ x => -( x + 4 )² + 31 ≤ 31

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

=> MaxA = 31 <=> x = -4

B = 4x - x² + 2

= -( x² - 4x + 4 ) + 6

= -( x - 2 )² + 6

-( x - 2 )² ≤ 0 ∀ x => -( x - 2 )² + 6 ≤ 6

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxB = 6 <=> x = 2

C = -x² - y² + 4x + 4y + 2 ( -x² mới ra :v )

= -( x² - 4x + 4 ) - ( y² - 4y + 4 ) + 10

= -( x - 2 )² - ( y - 2 )² + 10

\(\hept{\begin{cases}-\left(x-2\right)^2\le0\forall x\\-\left(y-2\right)^2\le0\forall y\end{cases}}\Rightarrow-\left(x-2\right)^2-\left(y-2\right)^2+10\le10\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

=> MaxC = 10 <=> x = y = 2

Answer 4

\(A=15-8x-x^2\)

\(=-\left(x^2+8x+15\right)\)

\(=-\left(x+4\right)^2+31\le31\forall x\)

Dấu"="xảy ra khi \(-\left(x+4\right)^2=0\Rightarrow x=-4\)

Vậy \(Max_A=31\Leftrightarrow x=-4\)

\(B=4x-x^2+2\)

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

\(=-\left(x-2\right)^2+6\le6\forall x\)

Dấu"="xảy ra khi \(-\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy \(Max_B=6\Leftrightarrow x=2\)