ĐKXĐ:

\(x^4-2x^2+3\ne0\)

\(\Leftrightarrow\left(x^2-1\right)^2+2\ne0\) (luôn đúng)

Hàm xác định trên R hay \(D=R\)

ĐKXĐ: \(x^2-x+1\ne0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ne0\) (luôn đúng)

Hàm số xác định với mọi x hay \(D=R\)

ĐKXĐ:

\(\left(1-x\right)\left(x^2-4x+3\right)\ne0\)

\(\Leftrightarrow-\left(x-1\right)^2\left(x-3\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)

Hay \(D=R\backslash\left\{1;3\right\}\)

TXĐ: D=(1;+\(\infty\))

Hàm số xác định: \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ne0\\x-1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x>1\end{matrix}\right.\) \(\Rightarrow x>1\)

Vậy \(D=\left(1;+\infty\right)\)

Ta có:

`@-1 <= sin x <= 1`

  `<=>0 <= 1+sin x <= 2=>1+sin x >= 0`

`@-1 <= cos x <= 1`

`<=>1 >= -cos x >= -1`

`<=>2 >= 1-cos x >= 0=>1-cos x >= 0`

Hàm số xác định `<=>[1+sin x]/[1-cos x] >= 0`

     `<=>{(1+sin x >= 0(L Đ)),(1-cos x > 0):}<=>1-cos x ne 0<=>x ne k2\pi (k in ZZ)`

   `=>TXĐ: D=R\\{k2\pi| k in ZZ}`.

ĐK: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\sin^4x-cos^4x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{\pi}{4}\ne k\pi\\\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\cos2x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\2x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

ĐKXĐ:

a. \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)  \(\Rightarrow D=[1;+\infty)\backslash\left\{3\right\}\)

b. \(D=R\)

c. \(x+3>0\Rightarrow x>-3\Rightarrow D=\left(-3;+\infty\right)\)

d. \(\left|x-2\right|\ge0\Rightarrow x\in R\Rightarrow D=R\)

\(y=\dfrac{cos3x}{1-sinx}+tanx=\dfrac{cos3x}{1-sinx}+\dfrac{sinx}{cosx}\)

Hàm số xác định khi \(\left\{{}\begin{matrix}1-sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sinx\ne1\\cosx\ne0\end{matrix}\right.\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)

Có: `-1 <= sin x <= 1`

`<=>-2 <= sin x-1 <= 0=>sin x-1 <= 0`

Để hàm số đã cho xác định `<=>sin x-1 >= 0`    Mà `sin x - 1 <= 0`

         `=>sin x -1=0<=>x=\pi/2+k2\pi`   `(k in ZZ)`

  `=>TXĐ: D=\pi/2 +k2\pi`   `(k in ZZ)`.