$8$) $H = - |x-8| - |9-x|$
$⇔ H = -(|x-8| + |9-x|)$
$⇔ -H = |x-8| + |9-x|$
Áp dụng BĐT $|a| + |b| ≥ |a+b|$ dấu : "$=$" xảy ra $⇔$ $ab ≥0$ ta có:
$|x-8| + |9-x| ≥ |(x-8)+(9-x)| = 1$
$⇔ -H ≥ 1$
$⇒$ $H ≤ -1$. Dấu "$=$" xảy ra $⇔$ $(x-8)(9-x) ≥ 0$
$⇔ (x-8)(x-9) ≤ 0$
Mà $x-8 > x-9 ∀ x$
$⇒$ $\left\{\begin{matrix} x-8≥0\\x-9 ≤0\end{matrix}\right.$
$⇔$ $8 ≤ x ≤ 9$
Vậy $H_{max}=-1$. Dấu "$=$" xảy ra khi $8 ≤ x ≤ 9$.
$9$) $M = -|2x-3| - |2x-5| + 8$
$⇔ M = -(|2x-3| + |2x-5| - 8)$
$⇔ -M = |3-2x| + |2x-5| - 8$
Áp dụng BĐT $|a| + |b| ≥ |a+b|$ dấu : "$=$" xảy ra $⇔$ $ab ≥0$ ta có:
$|3-2x| + |2x-5| - 8 ≥ |(3-2x)+(2x-5)| - 8 = -6$
$⇔ -M ≥ -6$
$⇔ M ≤ 6$. Dấu "$=$" xảy ra $⇔$ $(3-2x)(2x+5) ≥ 0$
$⇔ (2x-3)(2x+5) ≤ 0$
Mà $2x+5 > 2x-3 ∀ x$
$⇒$ $\left\{\begin{matrix} 2x+5≥0\\2x-3 ≤0\end{matrix}\right.$
$⇔$ $\dfrac{-5}{2} ≤ x ≤\dfrac{3}{2}$
Vậy $M_{max}=6$. Dấu "$=$" xảy ra khi $\dfrac{-5}{2} ≤ x ≤\dfrac{3}{2}$.
$10$) $N = -|3x+5| - |3x+7|$
$⇔ N = -(|3x+5| + |3x+7|)$
$⇔ -N = |-3x-5| + |3x+7|$
Áp dụng BĐT $|a| + |b| ≥ |a+b|$ dấu : "$=$" xảy ra $⇔$ $ab ≥0$ ta có:
$|-3x-5| + |3x+7| ≥ |(-3x-5)+(3x+7)| = 2$
$⇔ -N ≥ 2$
$⇔ N ≤ -2$. Dấu "$=$" xảy ra $⇔$ $(-3x-5)(3x+7) ≥ 0$
$⇔ (3x-5)(3x+7) ≤ 0$
Mà $3x+7 >3x-5 ∀ x$
$⇒$ $\left\{\begin{matrix} 3x+7≥0\\3x-5 ≤0\end{matrix}\right.$
$⇔$ $\dfrac{-7}{3} ≤ x ≤\dfrac{5}{3}$
Vậy $N_{max}=-2$. Dấu "$=$" xảy ra khi $\dfrac{-7}{3} ≤ x ≤\dfrac{5}{3}$ .
