\(32x^2-8x+11\)
\(=32\left(x^2-\frac{1}{4}x+\frac{11}{34}\right)\)
\(=32\left(x^2-2\cdot x\cdot\frac{1}{8}+\frac{1}{64}+\frac{335}{1088}\right)\)
\(=32\left[\left(x-\frac{1}{8}\right)^2+\frac{335}{1088}\right]\)
\(=32\left(x-\frac{1}{8}\right)^2+\frac{335}{34}\ge\frac{335}{34}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{8}\)
@ trần Thanh Phương@ em nhầm chỗ này rồi em:
\(A=32\left(x^2-\frac{1}{4}x+\frac{11}{32}\right)=32\left(x^2-2.x.\frac{1}{8}+\frac{1}{64}\right)+\frac{21}{2}=32\left(x-\frac{1}{8}\right)^2+\frac{21}{2}\)\(\ge\frac{21}{2}\)
"=" xảy ra khi x=1/8
Vậy min A=21/2 khi x=1/8
