Question

$2 {x }^{2} - 3x + 4 > 0$
$9 {x}^{2} - 6x + 1 \leqslant 0$
$4 {x}^{2} - 20x + 25 < 0$
Даю 30 баллов​

Answer 1

$1)\ \ 2x^2-3x+4>0\ \ ,\ \ \ D=9-32=-23<0\ \ \ \Rightaarrow \ \ x\in (-\infty ;+\infty )\\\\\\2)\ \ 9x^2-6x+1\leq 0\ \ \Rightarrow \ \ \ (3x-1)^2\leq 0\ \ \Rightarrow \ \ 3x-1=0\ \ ,\ \ x=\dfrac{1}{3}\\\\\\3)\ \ 4x^2-20x+25<0\ \ \ \Rightarrow \ \ \ (2x-5)^2<0\ \ \ \Rightarrow \ \ \ x\in \varnothing$