Bài 3:
a: Ta có: \(A=16x^2+8x+5\)
\(=16x^2+8x+1+4\)
\(=\left(4x+1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)
b: Ta có: \(B=2x^2-5x\)
\(=2\left(x^2-\dfrac{5}{2}x\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)
\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)
a: Ta có: \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^3+6x^2-12x+8-x^3-125=-108\)
\(\Leftrightarrow7x^2-6x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{6}{7}\end{matrix}\right.\)
