Question

2,3 (найти область определения)

Answer 1

$y(x)=sqrt{2x^2-5x-3}$

$2x^2-5x-3 geq 0\\ D=(-5)^2-4*2*(-3)=25+24=49=7^2\\ x_{1,2}=frac{-(-5)pmsqrt{7^2}}{2*2}=frac{5pm7}{4}\\ x_1=3 x_2=-frac{1}{2}\\ a(x-x_1)(x-x_2)=0\\ 2*(x-3)*(x-(-frac{1}{2})) geq 0\\ (x-3)*(x-(-0.5)) geq 0\\ +++++++[-0.5]------[3]+++++++++ textgreater x\\ xin(-infty; -0.5)cup(3; +infty)$