A.
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(|x+1|+|x-3|=|x+1|+|3-x|\geq |x+1+3-x|=4\)
\(\Rightarrow A=|x+1|+|x-3|+|2x-2|\geq 4+|2x-2|\geq 4+0=4\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x+1)(3-x)\geq 0\\ 2x-2=0\end{matrix}\right.\Rightarrow x=1\)
Vậy $A_{\min}=4$ khi $x=1$
B.
Vì \((x-2)^2\geq 0; (3y-1)^4\geq 0, \forall x,y\in\mathbb{R}\)
\(\Rightarrow B=2(x-2)^2+(3y-1)^4-5\geq 2.0+0-5=-5\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-2)^2=0\\ (3y-1)^4=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=\frac{1}{3}\end{matrix}\right.\)
Vậy $B_{\min}=-5$ khi $x=2; y=\frac{1}{3}$
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