Tìm TXĐ:
a) y=2cot(x- pi/3)/2cos x+căn3
b) y=-2sin2x/căn3 – tan(pi/4-x)
c) y=căn sinx+2 /cos x+ cos 2x
d)y=tan(2x-pi/3)/2sin^2 2x-1
e) y=2-tan4x/cosx.sin(3pi/2-2x)
f)y=căn 5+sin x-cos x
Tìm TXĐ:
a) y=2cot(x- pi/3)/2cos x+căn3
b) y=-2sin2x/căn3 – tan(pi/4-x)
c) y=căn sinx+2 /cos x+ cos 2x
d)y=tan(2x-pi/3)/2sin^2 2x-1
e) y=2-tan4x/cosx.sin(3pi/2-2x)
f)y=căn 5+sin x-cos x
Đáp án:
a) $D = R\backslash \left\{ {\dfrac{\pi }{3} + k\pi ;\dfrac{{5\pi }}{6} + k2\pi ; – \dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}$
b) $D = R\backslash \left\{ {\dfrac{{ – \pi }}{4} + k\pi ;\dfrac{{ – \pi }}{{12}} + k\pi |k \in Z} \right\}$
c) $D = R\backslash \left\{ {\pi + k2\pi ;\dfrac{\pi }{3} + k2\pi ;\dfrac{{ – \pi }}{3} + k2\pi |k \in Z} \right\}$
d) $D = R\backslash \left\{ {\dfrac{{5\pi }}{{12}} + k\dfrac{\pi }{2};\dfrac{\pi }{8} + k\dfrac{\pi }{4}|k \in Z} \right\}$
e) $D = R\backslash \left\{ {\dfrac{\pi }{8} + k\dfrac{\pi }{4};\dfrac{\pi }{2} + k\pi ;\dfrac{{3\pi }}{4} + k\dfrac{\pi }{2}|k \in Z} \right\}$
f) $D=R$
Giải thích các bước giải:
$a)y = \dfrac{{2\cot \left( {x – \dfrac{\pi }{3}} \right)}}{{2\cos x + \sqrt 3 }}$ có nghĩa
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\sin \left( {x – \dfrac{\pi }{3}} \right) \ne 0\\
\cos x \ne \dfrac{{ – \sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x – \dfrac{\pi }{3} \ne k\pi \\
x \ne \dfrac{{5\pi }}{6} + k2\pi \\
x \ne – \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{3} + k\pi \\
x \ne \dfrac{{5\pi }}{6} + k2\pi \\
x \ne – \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$ \Rightarrow D = R\backslash \left\{ {\dfrac{\pi }{3} + k\pi ;\dfrac{{5\pi }}{6} + k2\pi ; – \dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}$
$b)y = \dfrac{{ – 2\sin 2x}}{{\sqrt 3 – \tan \left( {\dfrac{\pi }{4} – x} \right)}}$ có nghĩa
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\cos \left( {\dfrac{\pi }{4} – x} \right) \ne 0\\
\tan \left( {\dfrac{\pi }{4} – x} \right) \ne \sqrt 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{\pi }{4} – x \ne \dfrac{\pi }{2} + k\pi \\
\dfrac{\pi }{4} – x \ne \dfrac{\pi }{3} + k\pi
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{{ – \pi }}{4} + k\pi \\
x \ne \dfrac{{ – \pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
\to D = R\backslash \left\{ {\dfrac{{ – \pi }}{4} + k\pi ;\dfrac{{ – \pi }}{{12}} + k\pi |k \in Z} \right\}
\end{array}$
$c)y = \dfrac{{\sqrt {\sin x + 2} }}{{\cos x + \cos 2x}}$ có nghĩa
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\sin x + 2 \ge 0\left( {ld} \right)\\
\cos x + \cos 2x \ne 0
\end{array} \right.\\
\Leftrightarrow \cos x + \cos 2x \ne 0\\
\Leftrightarrow 2{\cos ^2}x + \cos x – 1 \ne 0\\
\Leftrightarrow \left( {\cos x + 1} \right)\left( {2\cos x – 1} \right) \ne 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\cos x \ne – 1\\
\cos x \ne \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne \pi + k2\pi \\
x \ne \dfrac{\pi }{3} + k2\pi \\
x \ne \dfrac{{ – \pi }}{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$\to$ $D = R\backslash \left\{ {\pi + k2\pi ;\dfrac{\pi }{3} + k2\pi ;\dfrac{{ – \pi }}{3} + k2\pi |k \in Z} \right\}$
$d)y = \dfrac{{\tan \left( {2x – \dfrac{\pi }{3}} \right)}}{{2{{\sin }^2}2x – 1}} = \dfrac{{\tan \left( {2x – \dfrac{\pi }{3}} \right)}}{{ – \cos 4x}}$ có nghĩa
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\cos \left( {2x – \dfrac{\pi }{3}} \right) \ne 0\\
\cos 4x \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2x – \dfrac{\pi }{3} \ne \dfrac{\pi }{2} + k\pi \\
4x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{{5\pi }}{{12}} + k\dfrac{\pi }{2}\\
x \ne \dfrac{\pi }{8} + k\dfrac{\pi }{4}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$\to$ $D = R\backslash \left\{ {\dfrac{{5\pi }}{{12}} + k\dfrac{\pi }{2};\dfrac{\pi }{8} + k\dfrac{\pi }{4}|k \in Z} \right\}$
$e)y = \dfrac{{2 – \tan 4x}}{{\cos x.\sin \left( {\dfrac{{3\pi }}{2} – 2x} \right)}}$ có nghĩa
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\cos 4x \ne 0\\
\cos x \ne 0\\
\sin \left( {\dfrac{{3\pi }}{2} – 2x} \right) \ne 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{2} + k\pi \\
\dfrac{{3\pi }}{2} – 2x \ne k\pi
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{8} + k\dfrac{\pi }{4}\\
x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{{3\pi }}{4} + k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$\to$ $D = R\backslash \left\{ {\dfrac{\pi }{8} + k\dfrac{\pi }{4};\dfrac{\pi }{2} + k\pi ;\dfrac{{3\pi }}{4} + k\dfrac{\pi }{2}|k \in Z} \right\}$
$f)y = \sqrt 5 + \sin x – \cos x$ có nghĩa với mọi $x\in R$
$\to D=R$