Bài làm:
Ta có: \(\left|x-4\right|.\left(2-\left|x-4\right|\right)\)
\(=-\left|x-4\right|^2+2.\left|x-4\right|\)
\(=-\left(\left|x-4\right|^2-2.\left|x-4\right|+1\right)+1\)
\(=-\left(\left|x-4\right|-1\right)^2+1\le1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(-\left(\left|x-4\right|-1\right)^2=0\Leftrightarrow\left|x-4\right|=1\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
Vậy Max = 1 khi x = 3 hoặc x = 5
có cái bài dễ vậy ko làm dc
\(\left|x-4\right|\left(2-\left|x-4\right|\right)\)
+) Nếu \(x\ge4\)
\(\left|x-4\right|\left(2-\left|x-4\right|\right)=\left(x-4\right)\left(6-x\right)=6x-x^2-24+4x\)
\(=x^2-10x-24=-\left(x-5\right)^2+1\le1\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\) ( tmđk )
+) Nếu \(x< 4\)
\(\left|x-4\right|\left(2-\left|x-4\right|\right)=\left(-x+4\right)\left(2+x-4\right)=\left(4-x\right)\left(x-2\right)\)
\(=4x-8-x^2+2x=-x^2+6x-8=-\left(x-3\right)^2+1\le1\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\) ( tmđk )
Vậy GTLN của bt trên = 1 \(\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)
