$(x+1)(x+2)(x+3)(x+4)$ $=(x+1)(x+4)(x+2)(x+3)$ $=(x²+5x+4)(x²+5x+6)$ Đặt $x²+5x=t$ $A=(t+4)(t+6)$ $=t²+10t+24$ $=t²+10t+25-1$ $=(t+5)²-1$ Vì $(t+5)²≥0$ $→(t+5)²-1≥-1$ $→$ Dấu “=” xảy ra khi $t+5=0$ $↔x²+5x+5=0$ $↔x²+2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}=0$ $↔(x+\dfrac{5}{2})²=\dfrac{5}{4}$ \(\leftrightarrow\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{5}}{2}\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=\dfrac{\sqrt{5}-5}{2}\\x=\dfrac{-\sqrt{5}-5}{2}\end{array} \right.\) $→\min A=-1$ Bình luận
Đáp án: `min=-1` khi `x=(-5+-√5)/2` Giải thích các bước giải: `(x+1)(x+2)(x+3)(x+4)=[(x+1)(x+4)][(x+2)(x+3)]` `=(x^2+5x+4)(x^2+5x+6)` `=(x^2+5x+5-1)(x^2+5x+5+1)` `=(x^2+5x+5)^2-1>=-1` dấu = có khi `x^2+5x+5=0⇔x=(-5+-√5)/2` vậy `min=-1` khi `x=(-5+-√5)/2` Bình luận
$(x+1)(x+2)(x+3)(x+4)$
$=(x+1)(x+4)(x+2)(x+3)$
$=(x²+5x+4)(x²+5x+6)$
Đặt $x²+5x=t$
$A=(t+4)(t+6)$
$=t²+10t+24$
$=t²+10t+25-1$
$=(t+5)²-1$
Vì $(t+5)²≥0$
$→(t+5)²-1≥-1$
$→$ Dấu “=” xảy ra khi $t+5=0$
$↔x²+5x+5=0$
$↔x²+2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}=0$
$↔(x+\dfrac{5}{2})²=\dfrac{5}{4}$
\(\leftrightarrow\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{5}}{2}\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=\dfrac{\sqrt{5}-5}{2}\\x=\dfrac{-\sqrt{5}-5}{2}\end{array} \right.\)
$→\min A=-1$
Đáp án:
`min=-1` khi `x=(-5+-√5)/2`
Giải thích các bước giải:
`(x+1)(x+2)(x+3)(x+4)=[(x+1)(x+4)][(x+2)(x+3)]`
`=(x^2+5x+4)(x^2+5x+6)`
`=(x^2+5x+5-1)(x^2+5x+5+1)`
`=(x^2+5x+5)^2-1>=-1`
dấu = có khi `x^2+5x+5=0⇔x=(-5+-√5)/2`
vậy `min=-1` khi `x=(-5+-√5)/2`