1) \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\left(\left|\dfrac{3}{5}-x\right|\ge0,\forall x\in R\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{3}{5}-x=0\Leftrightarrow x=\dfrac{3}{5}\)
\(\Rightarrow GTNN\left(A\right)=\dfrac{1}{9}\left(khi.x=\dfrac{3}{5}\right)\)
2) \(A=\left|\dfrac{x}{3}-\dfrac{3}{2}\right|+\dfrac{2}{3}\ge\dfrac{2}{3}\left(\left|\dfrac{x}{3}-\dfrac{3}{2}\right|\ge0,\forall x\in R\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{x}{3}-\dfrac{3}{2}=0\Leftrightarrow x=3.\dfrac{3}{2}=\dfrac{9}{2}\)
\(\Rightarrow GTNN\left(A\right)=\dfrac{2}{3}\left(khi.x=\dfrac{9}{2}\right)\)
12) \(A=\dfrac{6}{11}+2\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\ge\dfrac{6}{11}\left(2\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\ge0,\forall x\in R\right)\)
Dấu "=" xảy ra khi \(\dfrac{5}{8}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{8}{5}.\dfrac{3}{5}=-\dfrac{24}{25}\)
\(\Rightarrow GTNN\left(A\right)=\dfrac{6}{11}\left(khi.x=-\dfrac{24}{25}\right)\)
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