a) \(4x^2+12x+10=\left(2x+3\right)^2+1\ge1\)
Dấu "="\(\Leftrightarrow x=-2\)
b) \(B=\left(3x-1\right)^2+4\ge4\)
Dấu "="\(\Leftrightarrow x=\frac{1}{3}\)
a, \(A=4x^2+12x+10\)
\(=\left(2x+1\right)^2+1\ge1\forall x\)
Dấu"=" xảy ra<=> \(\left(2x+1\right)^2=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(b,B=9x^2-6x+5\)
\(=\left(3x-1\right)^2+4\ge4\forall x\)
Dấu"=" xảy ra<=> \(\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\frac{1}{3}\)
a) \(B=4x^2+12x+9+1\)
B= \(\left(2x+6\right)^2+1\) > 1
Dấu '=' xảy ra khi \(2x+6=0\)
\(x=-3\)
b) B= \(9x^2-6x+1+4\)
= \(\left(3x-1\right)^2+4\) > 4
Dấu '=' xảy ra khi \(3x-1=0\)
\(x=\frac{1}{3}\)
Câu a) Dấu bằng khi x = \(\frac{-3}{2}\)nhé,nhầm