Question

1.Tìm Min:

B=4y²+4y+5

2.Tìm Max:

M:-x²-4x

Answer 1

\(B=4y^2+4y+5\)

\(=\left[\left(2y\right)^2+2.2y.1+1^2\right]+4\)

Vậy \(\left(2y+1\right)^2\ge0\)

\(\Rightarrow\left(2y+1\right)^2+4\ge4\)

Vậy GTNN là 4

Khi x = -1/2

Answer 2

1: \(B=4y^2+4y+5=\left(2y\right)^2+2\cdot y\cdot2+2^2+1=\left(2y+2\right)^2+1\)

Để B min 

Suy ra \(\left(2y+2\right)^2+1\)min

Mà \(\left(2y+2\right)^2\ge0\)

Suy ra \(\left(2y+2\right)^2+1\ge1\)

Vậy B min = 1

2: \(M=-x^2-4x=-x^2-2\cdot x\cdot2-4+4=-\left(x^2+2\cdot x\cdot2+2^2\right)+4=-\left(x+2\right)^2+4\)

Để M max

Suy ra \(-\left(x+2\right)^2+4\)max

Mà \(-\left(x+2\right)^2\le0\)

Suy ra\(-\left(x+2\right)^2+4\text{​​}\le4\)

Vậy M max = 4