Giải:
a) \(A=10+\left|\dfrac{1}{2}-x\right|\)
Vì \(\left|\dfrac{1}{2}-x\right|\ge0;\forall x\)
\(\Leftrightarrow10+\left|\dfrac{1}{2}-x\right|\ge10;\forall x\)
\(\Leftrightarrow A\ge10;\forall x\)
\(\Leftrightarrow A_{Min}=10\)
\("="\Leftrightarrow\dfrac{1}{2}-x=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b) \(B=\left|x+1,5\right|-5,7\)
\(\Leftrightarrow B=-5,7+\left|x+1,5\right|\)
Vì \(\left|x+1,5\right|\ge0;\forall x\)
\(\Leftrightarrow-5,7+\left|x+1,5\right|\ge-5,7;\forall x\)
\(\Leftrightarrow B\ge-5,7;\forall x\)
\(\Leftrightarrow B_{Min}=-5,7\)
\("="\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\)
Vậy ...
a) Ta có: \(A=10+\left|\dfrac{1}{2}-x\right|\ge10+0=10\)
Dấu "=" xảy ra \(\Leftrightarrow\left|\dfrac{1}{2}-x\right|=0\)
=> \(\dfrac{1}{2}-x=0\)
=> \(x=\dfrac{1}{2}\)
Vậy GTNN của A là 10 \(\Leftrightarrow x=\dfrac{1}{2}\)
b) Ta có: \(\left|x+1.5\right|-5.7=\left|x+5\right|-35\ge0-35=-35\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x+5\right|=0\)
=> \(x+5=0\)
=> \(x=-5\)
Vậy GTNN của B là -35 \(\Leftrightarrow x=-5\)
