A = 3 x | 1 - 2x | - 5
Ta co : | 1 - 2x | \(\ge\)0 nen 3 x | 1 - 2x | \(\ge\)0
A = 3 x | 1 - 2x | - 5 \(\ge\)- 5
Vậy min A = -5 \(\Leftrightarrow\)x = \(\frac{1}{2}\)
1 bài thôi . còn lại tương tự
bài cuối dùng BĐT : | a | + | b | \(\ge\)| a + b | nhé
max hả
ta có : | 2 - 5x | \(\ge\)0 nen -|2 - 5x | \(\le\)0
B = \(\frac{3}{4}-\left|2-5x\right|\le\frac{3}{4}\)
Vậy max B = \(\frac{3}{4}\)\(\Leftrightarrow x=\frac{2}{5}\)
D = | x - 17 | + | x - 12 |
D = | 17 - x | + | x - 12 | \(\ge\)| 17 - x + x - 12 | = 5
Dấu " = " xảy ra \(\Leftrightarrow\)( 17 - x ) ( x - 12 ) \(\ge\)0 hay 12 \(\le\)x \(\le\)17
Vậy min D = 5 \(\Leftrightarrow\)12 \(\le\)x \(\le\)17
\(C=\left(2x^2+1\right)^4-13\)
Ta có: \(2x^2\ge0\Rightarrow\left(2x^2+1\right)^4\ge1\)
\(\Rightarrow C\ge-12\)
\(\Rightarrow C_{min}=-12\Leftrightarrow x=0\)
\(B=\frac{3}{4}-\left|2-5x\right|\)
Ta có: \(\left|2-5x\right|\ge0\Rightarrow B\le\frac{3}{4}\)
\(\Rightarrow B_{max}=\frac{3}{4}\Leftrightarrow x=\frac{2}{5}\)