Bài này sử dụng tính chất cơ bản: \(\left|A\right|\pm A\ge0\) với mọi A
a.
\(A=\left|-x-3\right|+\left|4x+1\right|+\left|3x+5\right|+5x+2\)
\(A\ge\left|3x-2\right|+\left|3x+5\right|+5x+2=\left|3x-2\right|+\dfrac{3}{2}.\left|2x+\dfrac{10}{3}\right|+5x+2\)
\(A\ge\left|3x-2\right|+\left|2x+\dfrac{10}{3}\right|+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+5x+2\)
\(A\ge\left|5x+\dfrac{4}{3}\right|+5x+\dfrac{4}{3}+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+\dfrac{2}{3}\ge\dfrac{2}{3}\)
\(A_{min}=\dfrac{2}{3}\) khi \(2x+\dfrac{10}{3}=0\Rightarrow x=-\dfrac{5}{3}\)
b. Tương tự
\(B\ge\left|5x+7\right|+\left|x+\dfrac{5}{4}\right|+3\left|x+\dfrac{5}{4}\right|-6x+5\)
\(B\ge\left|6x+\dfrac{33}{4}\right|-\left(6x+\dfrac{33}{4}\right)+3\left|x+\dfrac{5}{4}\right|+\dfrac{53}{4}\ge\dfrac{53}{4}\)
\(B_{min}=\dfrac{53}{4}\) khi \(x=-\dfrac{5}{4}\)
Lời giải:
a. Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
\(A=|-x-3|+|4x+1|+|3x+5|+5x+2\)
\(\geq |-x-3+4x+1|+|3x+5|+5x+2=|3x-2|+|3x+5|+5x+2\)
Nếu $x\geq \frac{2}{3}$ thì:
$A\geq 3x-2+3x+5+5x+2=11x+5\geq 11.\frac{2}{3}+5=\frac{37}{3}$
Nếu $\frac{-5}{3}\leq x< \frac{2}{3}$ thì:
$A\geq 2-3x+3x+5+5x+2=9+5x\geq 9+5.\frac{-5}{3}=\frac{2}{3}$
Nếu $x< \frac{-5}{3}$ thì:
$A\geq 2-3x-3x-5+5x+2=-1-x>\frac{2}{3}$
Từ 3 TH trên suy ra $A_{\min}=\frac{2}{3}$ khi $x=\frac{-5}{3}$
b. Áp dụng BĐT $|a|+|b|\geq |a+b|$ thì:
\(B\geq |2x+3+3x+4|+|4x+5|-6x+5=|5x+7|+|4x+5|-6x+5\)
Nếu $x\geq \frac{-5}{4}$ thì:
$B\geq 5x+7+4x+5-6x+5=3x+17\geq 3.\frac{-5}{4}+17=\frac{53}{4}$
Nếu $\frac{-7}{5}\leq x< \frac{-5}{4}$ thì:
$B\geq 5x+7-4x-5-6x+5=-5x+7> -5.\frac{-5}{4}+7=\frac{53}{4}$
Nếu $x< \frac{-7}{5}$ thì:
$B\geq -5x-7-4x-5-6x+5=-15x-7> -15.\frac{-7}{5}-7=14$
Từ 3 TH trên suy ra $B_{\min}=\frac{53}{4}$. Giá trị này đạt tại $x=\frac{-5}{4}$