\(B=4x^2+4x+11\)
\(B=\left(2x+2\right)^2+7\)
\(\left(2x+2\right)^2\ge0\Rightarrow B\ge7\)
Dau "=" xảy ra khi 2x + 2 =0
<=> 2x = -2
<=> x = -1
Vậy Min B = 7 khi x =-1
\(B=4x^2+4x+11=4\left(x^2+x+\frac{11}{4}\right)\)
\(=4\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{10}{4}\right)\)
\(=4\left(x+\frac{1}{2}\right)^2+10\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\left(\forall x\right)\Rightarrow4\left(x+\frac{1}{2}\right)^2\ge0\)
Suy ra \(4\left(x+\frac{1}{2}\right)^2+10\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy BMin = 10 khi x = -1/2
\(B=4x^2+4x+11=4x^2+4x+1+10\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\)
\(\left(2x+1\right)^2+10\ge10\forall x\)
Hay: \(B\ge10\forall x\)
Dấu = xảy ra khi: \(\left(2x+1\right)^2=0\Rightarrow x=-\frac{1}{2}\)
=.= hok tốt!!
Ta có: \(B=4x^2+4x+11\)
\(=4x^2+2x+2x+11\)
\(=2x\left(2x+1\right)+2x+1+10\)
\(=2x\left(2x+1\right)+1\left(2x+1\right)+10\)
\(=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi \(\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)
Vậy giá trị nhỏ nhất của biểu thức B là 10 khi \(x=-\frac{1}{2}\).
