[tex]a. \frac{3+|x|+2}{4|x|-5}[/tex]
[tex]b. 5+\frac{15}{4|3x+7|+3}[/tex]
[tex]c. \frac{|2y+7|+13}{2|2y+7|+6}[/tex]
[tex]d. F = 4-|5x-2|- |3y+12|[/tex]
d.
$|5x-2|\ge 0;|3y+12|\ge 0\\\Rightarrow -|5x-2|\le 0;-|3y+12|\le 0\\\Rightarrow F=...\le 4$
Dấu "$=$" xảy ra khi $5x-2=0;3y+12=0 \Rightarrow x,y$
Vậy ...
c.
Ta có:
$2.\dfrac{|2y+7|+13}{2|2y+7|+6}=\dfrac{2|2y+7|+26}{2|2y+7|+6}=\dfrac{2|2y+7|+6+20}{2|2y+7|+6}=1+\dfrac{20}{2|2y+7|+6}$
$|2y+7|\ge 0\\\Rightarrow 2|2y+7|\ge 0\\\Rightarrow 2|2y+7|+6\ge 6\\\Rightarrow \dfrac{20}{2|2y+7|+6} \le \dfrac{20}{6}=\dfrac{10}{3}\\\Rightarrow1+\dfrac{20}{2|2y+7|+6} \le \dfrac{13}{3}\\\Rightarrow 2.\dfrac{|2y+7|+13}{2|2y+7|+6}\le \dfrac{13}{3}\\\Rightarrow \dfrac{|2y+7|+13}{2|2y+7|+6}\le \dfrac{13}{6}$
Dấu "$=$" xảy ra khi $2y+7=0\Rightarrow ...$
Vậy ...
b.
$|3x+7|\ge 0\\\Rightarrow 4|3x+7|\ge 0\\\Rightarrow 4|3x+7|+3\ge 3\\\Rightarrow \dfrac{15}{4|3x+7|+3}\le \dfrac{15}{3}=5\\\Rightarrow 5+\dfrac{15}{4|3x+7|+3}\le 5+5=10$
Dấu "$=$" xảy ra khi $3x+7=0\Rightarrow ...$
Vậy ...
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