Question

tìm tập xác định của hàm số y=1/[tan(2x-pi/6)-1]

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}2x-\dfrac{\Omega}{6}\ne\dfrac{\Omega}{2}+k\Omega\\tan\left(2x-\dfrac{\Omega}{6}\right)\ne1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x\ne\dfrac{4}{6}\Omega+k\Omega\\2x-\dfrac{\Omega}{6}\ne\dfrac{\Omega}{4}+k\Omega\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\ne\dfrac{1}{3}\Omega+\dfrac{k\Omega}{2}\\x\ne\dfrac{5}{24}\Omega+\dfrac{k\Omega}{2}\end{matrix}\right.\)

vậy: TXĐ là \(D=R\backslash\left\{\dfrac{1}{3}\Omega+\dfrac{k\Omega}{2};\dfrac{5}{24}\Omega+\dfrac{k\Omega}{2}\right\}\)