Question

ПОМОГИТЕ
СРОЧНО
30 БАЛЛОВ
ВО ВЛОЖЕНИИ

Answer 1

$\sqrt{x-1}+\sqrt{2x-1}=4-3x-2\sqrt{2x^2-3x+1} \ \ ,\\\\\\ODZ:\ a)\ 2x^2-3x+1\geq 0\ ,\ \ D=9-8=1\ ,\ \ x_1=\dfrac{1}{2}\ ,\ x_2=1\\\\2(x-\dfrac{1}{2})(x-1)\geq 0\ \ ,\ \ (2x-1)(x-1)\geq 0\ \ ,\\\\x\in (-\infty \, ;\, \dfrac{1}{2}\ ]\cup [\ 1\, ;+\infty \, )\\\\b)\ \ x-1\geq 0\ \ ,\ \ \ x\geq 1\\\\c)\ \ 2x-1\geq 0\ \ ,\ \ x\geq \dfrac{1}{2}\\\\\underline {\ x\in [\ 1\ ;+\infty \, )\ }$

$\sqrt{x-1}+\sqrt{2x-1}+2\sqrt{(2x-1)(x-1)}+3x-4=0\\\\\\\star \ \ 3x-2+2\sqrt{(2x-1)(x-1)}=\underbrace{(x-1)+(2x-1)}_{3x-2}+2\cdot \sqrt{2x-1}\cdot \sqrt{x-1}=\\\\=(\sqrt{2x-1}+\sqrt{x-1})^2\ \ \star$

$(\sqrt{x-1}+\sqrt{2x-1})+(\sqrt{2x-1}+\sqrt{x-1})^2-2=0\\\\t=\sqrt{x-1}+\sqrt{2x-1}\geq 0\ \ (!!!)\ \ ,\\\\t^2+t-2=0\ ,\ \ t_1=1\ ,\ t_2=-2<0\ \ ne\ podxodit\ \ (teorema\ Vieta)\\\\\\\sqrt{x-1}+\sqrt{2x-1}=1\ \ \Rightarrow \ \ \sqrt{x-1}=1-\sqrt{2x-1}\ \ ,\\\\x-1=1-2\sqrt{2x-1}+(2x-1)\\\\2\sqrt{2x-1}=x+1\ \ ,\\\\4\, (2x-1)=x^2+2x+1\ \ ,\ \ \ 8x-4=x^2+2x+1\ \ ,\ \ \ x^2-6x+5=0\ ,\\\\x_1=1\in ODZ\ ,\ x_2=5\in ODZ\ \ \ (teorema\ Vieta)$

$Proverka:\ x=1:\ \sqrt{1-1}+\sqrt{2-1}=4-3-2\sqrt{0}\ ,\ \ 1=1\ ,\ verno\\\\x=5:\ \sqrt{5-1}+\sqrt{10-1}=4-15-2\sqrt{50-15+1}\ ,\ \ 2+3\ne -11-12\\\\\\Otvet:\ \ x=1\ .$