Question

√(10x^2-11x-6)^-1 помогите найти область определения

Answer 1

$y = sqrt{dfrac{1}{10x^{2}-11x-6}} \ \ left{begin{gathered} 10x^{2}-11x-6 ne 0 (a) \ \ sqrt{dfrac{1}{10x^{2}-11x-6}} ge 0 (b) end{gathered} right.$

$(a): 10x^{2} - 11x - 6 ne 0 \ D = 121 + 240 = 361 \ \ x_{1} = dfrac{11 + 19}{20} ne dfrac{3}{2} ; x_{2} ne -dfrac{2}{5} \ \ (b): sqrt{dfrac{1}{10x^{2}-11x-6}} ge 0 \ \ dfrac{1}{10x^{2}-11x-6} ge 0 \ \ dfrac{1}{(x-dfrac{3}{2})(x+dfrac{2}{5})} ge 0 \ \ x in (-infty ; -dfrac{2}{5})cup (dfrac{3}{2}; +infty)$

$D(y) = (-infty ; -dfrac{2}{5})cup (dfrac{3}{2}; +infty)$

Ответ: D(y) = (-∞; -2/5)∪(3/2; +∞)

Answer 2

Умненько!