a, để Amax khi\(\dfrac{15}{4\left|3x+7\right|+3}max\) khi:
\(\left\{{}\begin{matrix}4\left|3x+7\right|+3min\\4\left|3x+7\right|+3>0\end{matrix}\right.\)
mà\(4\left|3x+7\right|+3\ge3\)nên max A=10 khi x=\(\dfrac{-7}{3}\)
a: 4|3x+7|+3>=3
=>15/4|3x+7|+3<=5
=>A<=10
Dấu = xảy ra khi x=-7/3
b: 8|15x-21|+7>=7
=>21/8|15x-21|+7<=3
=>B<=3-1/3=8/3
Dấu = xảy ra khi x=7/5