Question

Tìm GTLN của biểu thức

A = 3 - ( x + 2 )^2

B = -( x - 2 )^2 + ( y + 1 )^2 - 9

Answer 1

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

\(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow3-\left(x+2\right)^2\le3\)

Dấu "=" xảy ra khi:

\(\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)

\(\Rightarrow MAX_A=3\) khi \(x=-2\)

\(B=-\left(x-2\right)^2-\left(y+1\right)^2-9\) (đã sửa đề)

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\left(x-2\right)^2\le0\\-\left(y+1\right)^2\le0\end{matrix}\right.\)

\(\Rightarrow-\left(x-2\right)^2-\left(y+1\right)^2\le0\)

\(\Rightarrow-\left(x-2\right)^2-\left(y+1\right)^2-9\le9\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}-\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\\-\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\end{matrix}\right.\)

\(\Rightarrow MAX_B=9\) khi \(x=2;y=-1\)

\(\)

Answer 2

a. Vì \(\left(x+2\right)^2\ge0\forall x\Rightarrow-\left(x+2\right)^2\le0\)

=> \(3- \left(x+2\right)^2\le3\)

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

Vậy \(A_{MAX}=3\) khi x = -2

b. Đề sai