3x+2-2[tex]\sqrt{x+3}[/tex]*[tex]\sqrt{2x-1}[/tex]=[tex]\sqrt{x+3}[/tex]-[tex]\sqrt{2x-1}[/tex]. Created by Booguboo. matematika-ru

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3x+2-2[tex]\sqrt{x+3}[/tex]*[tex]\sqrt{2x-1}[/tex]=[tex]\sqrt{x+3}[/tex]-[tex]\sqrt{2x-1}[/tex]

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pavlikleon

$3x+2-2\sqrt{x+3} \sqrt{2x-1}=x+3-2\sqrt{x+3} \sqrt{2x-1}+2x-1=(\sqrt{x+3})^{2}-2\sqrt{x+3} \sqrt{2x-1}+(\sqrt{2x-1})^{2}=(\sqrt{x+3}- \sqrt{2x-1})^{2}\\ y=\sqrt{x+3}- \sqrt{2x-1}\\\\y^{2}=y\\y_{1}=0, y_{2}=1\\y_{1})\\\sqrt{x+3}-\sqrt{2x-1}=0\\\sqrt{x+3}=\sqrt{2x-1}\\x+3=2x-1\\x_{1}=4\\y_{2})\\y_{2}^{2}=3x+2-2\sqrt{x+3} \sqrt{2x-1}=1\\2\sqrt{x+3} \sqrt{2x-1}=3x+1\\4(x+3)(2x-1)=(3x+1)^{2}\\8x^{2}+20x-12=9x^{2}+6x+1\\x^{2}-14x+13=0\\x_{2}=1, x_{3}=13$

Все три полученные ответа входят в ОДЗ ( x≥0.5)

Ответ: х∈{1; 4; 13}

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