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/

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 $x\in R$

$\to D=R$