Đặt \(A=\left(x+1\right)^2+\left(3x-4\right)^2\)
\(\Rightarrow A=x^2+2x+1+9x^2-24x+16\)
\(A=10x^2-22x+17=10\left(x-\dfrac{11}{10}\right)^2+\dfrac{49}{10}\ge\dfrac{49}{10}\)
\(A_{min}=\dfrac{49}{10}\) khi \(x=\dfrac{11}{10}\)
Ta có: \(\left(x+1\right)^2+\left(3x-4\right)^2\)
\(=x^2+2x+1+9x^2-24x+16\)
\(=10x^2+22x+17\)
\(=10\left(x^2+\dfrac{11}{5}x+\dfrac{17}{10}\right)\)
\(=10\left(x^2+2\cdot x\cdot\dfrac{11}{10}+\dfrac{121}{100}+\dfrac{49}{100}\right)\)
\(=10\left(x+\dfrac{11}{10}\right)^2+\dfrac{49}{10}\ge\dfrac{49}{10}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{11}{10}\)
