Ta có :
\(\left|x-1\right|=\left|1-x\right|\)
\(\Leftrightarrow B=\left|x+1\right|+\left|3x-4\right|+\left|1-x\right|+5\)
Lại có :
\(\left|x+1\right|+\left|1-x\right|\ge\left|x+1+1-x\right|=2\)
\(\Leftrightarrow B=\left|x+1\right|+\left|3x-4\right|+\left|1-x\right|+5\ge\left|3x-4\right|+2+5\)
\(\Leftrightarrow B\ge\left|3x-4\right|+7\)
Với mọi x ta có :
\(\left|3x-4\right|\ge0\)
\(\Leftrightarrow\left|3x-4\right|+7\ge7\)
\(\Leftrightarrow B\ge7\)
Dấu "=" xảy ra \(\Leftrightarrow\left|3x-4\right|=0\Leftrightarrow x=\frac{4}{3}\)
Vậy...
Đúng 0
Bình luận (1)
