Question

Tìm GTNN biểu thức

a, A = ( x + 1 )^2 - 5

b, B = ( x -1 )^2 + ( y + 1 )^2 + 3

Answer 1

a. Ta có: \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2-5\ge-5\)

Dấu ''='' xảy ra khi \(\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy \(A_{MIN}=-5\) khi x = -1

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

=> \(\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

=> \(\left(x-1\right)^2+\left(y+1\right)^2+3\ge3\)

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

Vậy \(B_{MIN}=3\) khi x = 1 và y = -1