Question

1. |2x²-5x|-3=0
2. |x²-7x +5|=1
​

Answer 1

$1)\ \ |2x^2-5x|-3=0\\\\a)\ \ 2x^2-5x\geq 0\ \ \to \ \ x(2x-5)\geq 0\ \ ,\ \ x\in (-\infty ;0\, ]\cup [\, 2,5\ ;+\infty \, )\\\\2x^2-5x-3=0\ \ ,\ \ x_1=-0,5\ \ ,\ \ x_2=3\\\\b)\ \ 2x^2-5x$

$2)\ \ |x^2-7x+5|=1\\\\a)\ \ x^2-7x+5\geq 0\ \ \to \ \ \ x\in \Big(-\infty ;\dfrac{7-\sqrt{29}}{2}\ \Big]\cup \Big[\ \dfrac{7+\sqrt{29}}{2}\ ;+\infty \, \Big)\\\\x\in (-\infty \, ;\, 0,81\ ]\cup [\ 6,19\ ;+\infty \, )\\\\x^2-7x+5=1\ \ ,\ \ x^2-7x+4=0\ ,\\\\D=33\ \ ,\ \ x_{1,2}=\dfrac{7\pm \sqrt{33}}{2}\ \ ,\ \ x_1\approx 0,63\ ,\ x_2\approx 6,37\\\\b)\ \ x^2-7x+5$

$Otvet:\ \ x_1=\dfrac{7-\sqrt{33}}{2}\ ,\ x_2=\dfrac{7+\sqrt{33}}{2}\ ,\ x_3=1\ ,\ x_4=6\ .$