Đáp án:
$GTLN$ $của$ $y$ $là$ $\frac{-3}{4}$ $tại$ $x = \frac{-1}{4}$
Giải thích các bước giải:
$với$ $-1<x<1$ $ta$ $có:$
$y = - 4(x² - x + 1) + 3|2x - 1|$
$→y = - (4x² - 4x + 1) - 3 + 3|2x - 1| $
$→ y = - (2x - 1)² + 3|2x - 1| -3 $
$→ y = - (|2x -1|)² + 3|2x - 1|-3$
$→ y = - [(|2x - 1|)² - 3|2x - 1|)-3$
$→ y = - [(|2x-1|)² - 2.|2x-1|.\frac{3}{2} + \frac{9}{4}] - 3 + \frac{9}{4}$
$→y = - (|2x-1| - \frac{3}{2})² - \frac{3}{4}$
$vì$ $(|2x - 1| - \frac{3}{2})² \geq 0$
$→ - (|2x - 1| - \frac{3}{2})² \leq 0$
$→ - (|2x - 1| - \frac{3}{2})² - \frac{3}{4} \leq \frac{-3}{4}$
$→ y \leq \frac{-3}{4}$
$Dấu$ $"="$ $xảy$ $ra$ $→ |2x - 1| - \frac{3}{2} = 0$ $→ |2x - 1| = \frac{3}{2}$
$→\left[ \begin{array}{l}2x - 1 = \frac{3}{2}\\2x - 1 = \frac{-3}{2}\end{array} \right.→\left[ \begin{array}{l}2x = \frac{5}{2}\\2x = \frac{-1}{2}\end{array} \right.→\left[ \begin{array}{l}x= \frac{5}{4}(k^{0} tmđk)\\x=\frac{-1}{4}(tmđk)\end{array} \right.$
$Vậy$ $GTLN$ $của$ $y$ $là$ $\frac{-3}{4}$ $tại$ $x = \frac{-1}{4}$
