\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x+6\right)\left(x-1\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-6^2.P_{min}\Leftrightarrow x^2+5xđạtGTNN\)
\(x^2+5x\ge0\Leftrightarrow x\left(x+5\right)\ge0\)
Dấu "=" xảy ra <=> \(x\in\left\{0;-5\right\}\)
Vậy: Pmin=-36 <=> x E {0;-5}
\(\frac{3x^2+2x+3}{x^2+1}=3+\frac{2x}{x^2+1}\Rightarrow Q_{min}\Leftrightarrow\frac{2x}{x^2+1}đạtGTNN\)
\(Matkhac:\frac{2x}{x^2+1}\ge0\)
Dấu "=" xảy ra <=> x=0 \(Vậy:Q_{min}=2\Leftrightarrow x=0\)
P/s: ms hok lp 6 sai sót mong mn thông cảm
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=\left[x^2+5x-6\right]\left[x^2+5x+6\right]\)
Đặt \(x^2+5x=a\)
\(\Rightarrow P=\left(a-6\right)\left(a+6\right)=a^2-36\ge-36\)
Vậy \(P_{min}=-36\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
shitboChắc gì \(\frac{2x}{x^2+1}\ge0\),lỡ x âm ?
-_-
\(Q=2+\frac{a^2+2a+1}{a^2+1}=2+\frac{\left(a+1\right)^2}{a^2+1}\Rightarrow Q_{min}=2\)