1, Tìm GTNN:
a, C=x(x-3)(x-4)(x-7)
b, A=2/(6x-5-9x^2)
c, A= (3x^2-8x+6)/(x^2-2x+1)
2, Tìm GTNN + GTLN:
A= (3-4x)/(x^2+1)
$1.\\c, A= \dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{x^2-4x+4+2(x^2-2x+1)}{x^2-2x+1}\\=\dfrac{(x-2)^2}{(x-1)^2}+2$
Ta có : $(x-2)^2\geq 0\Leftrightarrow \dfrac{(x-2)^2}{(x-1)^2}\geq 0$
$\Leftrightarrow \dfrac{(x-2)^2}{(x-1)^2}+2\geq 2$
Dấu "=" xảy ra $\Leftrightarrow x=2$
Vậy Min A=2 $\Leftrightarrow x=2$
$2.$
*Tìm GTNN:
$A=\dfrac{3-4x}{x^2+1}\\=\dfrac{x^2-4x+4-x^2-1}{x^2+1}\\=\dfrac{(x^2-4x+4)-(x^2+1)}{x^2+1}\\=\dfrac{(x-2)^2}{x^2+1}-1$
Ta có: $(x-2)^2\geq 0\Leftrightarrow \dfrac{(x-2)^2}{x^2+1}\geq 0\Leftrightarrow \dfrac{(x-2)^2}{x^2+1}-1\geq -1$
Dấu "=" xảy ra $\Leftrightarrow x=2$
Vậy Min A=-1 $\Leftrightarrow x=2$
*Tìm GTLN:
$A=\dfrac{3-4x}{x^2+1}\\=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}\\=\dfrac{4(x^2+1)-(4x^2+4x+1)}{x^2+1}\\=4-\dfrac{(2x+1)^2}{x^2+1}$
Ta có: $(2x+1)^2\geq 0\Leftrightarrow \dfrac{(2x+1)^2}{x^2+1}\geq 0\Leftrightarrow 4-\dfrac{(2x+1)^2}{x^2+1}\leq 4$
Dấu "=" xảy ra $\Leftrightarrow x=\dfrac{-1}{2}$
Vậy Max A=4 $\Leftrightarrow x=\dfrac{-1}{2}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
