\(B=4y^2+4y+5\)
\(=\left[\left(2y\right)^2+2.2y.1+1^2\right]+4\)
Vậy \(\left(2y+1\right)^2\ge0\)
\(\Rightarrow\left(2y+1\right)^2+4\ge4\)
Vậy GTNN là 4
Khi x = -1/2
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
