\((x-1).(x+2).(x+3).(x+6) \)
\(=[(x-1).(x+6)].[(x+2).(x+3)] \)
\(=(x^2+5x-6).(x^2+5x+6) \)
\(=(x^2+5x)^2-36\) \(\ge-36\)
\(\Rightarrow\) Min \(= -36\)
\(\Leftrightarrow\) \(x=0\) hoặc \(x=-5\)
\(M=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
KL: \(Min_M=-36\) tại x=0
(x - 1)(x + 2)(x + 3)(x + 6)
= (x - 1)(x + 6)(x + 2)(x + 3)
= (x2 + 5x - 6)(x2 + 5x + 6)
đặt x2 + 5x = t
=> (t - 6)(t + 6) = t2 - 36
Ta có:
t2 >= 0 suy ra t2 - 36 >= -36
Vậy min = - 36
Dấu "=" xảy ra chỉ khi t2 = 0 chỉ khi x2 + 5x = 0
<=> x = 0 hoặc x = -5
làm ẩu quá, đọc lời giải kia mới biết mình thiếu x=-5 T_T!!!
M=\(\left(x-1\right)\left(x+6\right)\left(x+3\right)\left(x+2\right)\)
=\(\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x=a\)thì thay vào M :
M=\(\left(a-6\right)\left(a+6\right)=a^2-36\)
Do \(a^2\ge0\)(\(\forall a\))\(\Rightarrow\)\(a^2-36\ge-36\left(\forall a\right)\)
Vậy MinA = -36\(\Leftrightarrow a^2=0\Leftrightarrow a=0\)
Hay \(x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
