Đặt:
\(X=1,7+\left|3,4-x\right|\)
\(\left|3,4-x\right|\ge0\)
\(X_{MIN}\Rightarrow\left|3,4-x\right|_{MIN}\)
\(\left|3,4-x\right|_{MIN}=0\)
\(X_{MIN}=1,7+0=1,7\)
\(S=\left|x+2,8\right|-3,5\)
\(\left|x+2,8\right|\ge0\)
\(S_{MIN}\Rightarrow\left|x+2,8\right|_{MIN}\)
\(\left|x+2,8\right|_{MIN}=0\)
\(S_{MIN}=0-3,5=-3,5\)
a)Đặt 1.7+|3.4−x|=A
\(Do\left|3,4-x\right|\ge0\forall x\)
\(\Rightarrow1,7+\left|3,4-x\right|\ge1,7\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow\left|3,4-x\right|=0\)
\(\Rightarrow x=3,4\)
Vậy GTNN của A=1,7 \(\Leftrightarrow x=3,4\)
b) Đặt |x+2.8|−3.5=B
\(Do\left|x+2,8\right|\ge0\forall x\)
\(\Rightarrow\left|x+2,8\right|-3,5\ge-3,5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x+2,8\right|=0\)
\(\Rightarrow x=-2,8\)
Vậy GTNN của B =-3,5 \(\Leftrightarrow x=-2,8\)
