Question

Найдите сумму целых значений аргумента, принадлежащих области определения функции f(x)=( 1+ (2x+11/x^2-6x-7) )^5/6 - ( под корнем 81-x^2)

Answer 1

$f(x)=Big(1+dfrac{2x+11}{x^2-6x-7}Big)^frac 56-sqrt{81-x^2}\\\displaystyleleft { {{1+dfrac{2x+11}{x^2-6x-7}geq 0} atop {81-x^2}geq 0} right.~~Leftrightarrow~~left { {{dfrac{x^2-6x-7+2x+11}{x^2-6x-7}geq 0} atop {(9-x)(9+x)}geq 0} right.\\\left { {{dfrac{x^2-4x+4}{x^2-6x-7}geq 0} atop {(9-x)(9+x)}geq 0} right.~~Leftrightarrow~~left { {{dfrac{(x-2)^2}{(x-7)(x+1)}geq 0} atop {(9-x)(9+x)}geq 0} right.$

$displaystyle(x-2)^2geq 0;~xin R;~~~Rightarrow\\left { {{dfrac{(x-2)^2}{(x-7)(x+1)}geq 0} atop {(9-x)(9+x)}geq 0} right.~~~Leftrightarrow~~left { {{(x-7)(x+1)}> 0} atop {(9-x)(9+x)}geq 0} right.\\\ left { {{++++++(-1)----(7)++++++>x} atop {---[-9]+++++++++[9]---->x}} right. \\boldsymbol{D(f)= [-9;-1)cup (7;9]}$

-9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 + 8 + 9 = -27

Ответ : -27