\(A=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+\sqrt{13}\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+\sqrt{13}\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+\sqrt{13}\)
Đặt \(t=x^2+3x+1\)
\(=>A=\left(t-1\right)\left(t+1\right)+\sqrt{13}\)
\(=t^2-1+\sqrt{13}\)
\(=\left(x^2+3x+1\right)^2-1+\sqrt{13}\ge-1+\sqrt{13}\)
Vậy ....................
Câu B thôi buồn ngủ qué ko làm được , xin lỗi bạn .
A=\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+\sqrt{13}\)
= \(\left[x\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]\)
= \(\left(x^2+3x\right)\left(x^2+3x+2\right)+\sqrt{13}\)
Đặt t=\(x^2+3x+1\)
A= \(\left(t-1\right)\left(t+1\right)+\sqrt{13}\)
= \(t^2-1+\sqrt{13}\) \(\geq\) \(\sqrt{13}-1\)
Dấu = xảy ra khi \(t^2\)=0 \(\Leftrightarrow\) t=0 \(\Leftrightarrow\) \(x^2+3x+1=0\) \(\Leftrightarrow\) \(\left[\begin{array}{} x=\dfrac{-3+\sqrt{5}}2\\ x=\dfrac{-3-\sqrt{5}}2 \end{array} \right.\)(cái này bấm máy là ra)
Vậy MinA=\(\sqrt{13}-1\) khi \(\left[\begin{array}{} x=\dfrac{-3+\sqrt{5}}2\\ x=\dfrac{-3-\sqrt{5}}2 \end{array} \right.\)
b) B= \(\left(x^2-4x-5\right)\left(x^2-12x+27\right)-\sqrt{3}\)
= \(\left(x^2+x-5x-5\right)\left(x^2-3x-9x+27\right)-\sqrt{3}\)
= \(\left[x\left(x+1\right)-5\left(x+1\right)\right].\left[x\left(x-3\right)-9\left(x-3\right)\right]-\sqrt{3}\)
= \(\left(x+1\right)\left(x-5\right)\left(x-3\right)\left(x-9\right)-\sqrt{3}\)
= \(\left[\left(x+1\right)\left(x-9\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]-\sqrt{3}\)
= \(\left(x^2-8x-9\right)\left(x^2-8x+15\right)-\sqrt{3}\)
Đặt t=\(x^2-8x+3\)
\(\Rightarrow\) B= \(\left(t-12\right)\left(t+12\right)-\sqrt{3}\)
= \(t^2-144+\sqrt{3}\) \(\geq\) \(-144+\sqrt{3}\)
Dấu = xảy ra khi \(t^2=0\) \(\Leftrightarrow\) t=0 \(\Leftrightarrow\) \(x^2-8x+3\) =0 \(\Leftrightarrow\) \(\left[\begin{array}{} x=4+\sqrt{13}\\ x=4-\sqrt{13} \end{array} \right.\)(bạn chịu khó bấm máy,cái này dùng delta là ra)
Vậy MinB=\(-144+\sqrt{3}\) khi \(\left[\begin{array}{} x=4+\sqrt{13}\\ x=4-\sqrt{13} \end{array} \right.\)
\(A=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+\sqrt{13}\\ A=\left(x^2+3x\right)\left(x^2+3x+2\right)+\sqrt{13}\)
đặt: \(t=x^2+3x+1\), khi đó:
\(A=\left(t-1\right)\left(t+1\right)+\sqrt{13}\\ A=t^2-1+\sqrt{13}\)
vì: \(t^2\ge0\)
nên: \(A\ge\sqrt{13}-1\)
dấu "=" xảy ra khi t=0 \(\Rightarrow x^2+3x+1=0\)
\(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}=-1+\dfrac{9}{4}=\dfrac{5}{4}\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{5}{4}\\x+\dfrac{3}{2}=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{11}{4}\end{matrix}\right.\)
vậy GTNN của A là \(\sqrt{13}-1\) khi \(x=-\dfrac{1}{4}\) và \(x=-\dfrac{11}{4}\)
