Question

3x+2-2$sqrt{x+3}$*$sqrt{2x-1}$=$sqrt{x+3}$-$sqrt{2x-1}$

Answer 1

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

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

Ответ: х∈{1; 4; 13}