Bài 1: Tìm TXĐ các hàm số a) y = √(4x-3) + √(x+1) b) y = $\frac{√(x+5)-1}{√(x-1)}$ c) y = √(4-x) + √(5+x) – $\frac{1}{x}$ d) y = $\frac{√(3x-4) –

`a)` 

\(\left\{ \begin{array}{l}4x – 3 ≥ 0\\x + 1 ≥ 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x ≥ \dfrac{3}{4}\\x ≥ -1\end{array} \right.\) 

`=> D = [3/4; +∞)`

`b)` 

\(\left\{ \begin{array}{l}x + 5 ≥ 0\\x – 1 > 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x ≥ -5\\x > 1\end{array} \right.\) 

`=> D = (1; +∞)`

`c)`

\(\left\{ \begin{array}{l}4 – x ≥ 0\\5 + x ≥ 0\\x \ne 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x ≤ 4\\x ≥ -5\\x \ne 0\end{array} \right.\) 

`=> D = [-5; 4] \\ {0}`

`d)`

\(\left\{ \begin{array}{l}3x – 4 ≥ 0\\x + 1 ≥ 0\\2x – 1 > 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x ≥ \dfrac{4}{3}\\x ≥ -1\\x > \dfrac{1}{2}\end{array} \right.\)  

`=> D = [4/3; +∞)`

`e)`

\(\left\{ \begin{array}{l}4 – x ≥ 0\\4 + x ≥ 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x ≤ 4\\x ≥ -4\end{array} \right.\) 

`=> D = [-4; 4]`

`f)`

\(\left\{ \begin{array}{l}x – 1 > 0\\x + 1 > 0\\x + 3 ≥ 0\end{array} \right.\) 

`<=>` \(\left\{ \begin{array}{l}x > 1\\x > -1\\x ≥ -3\end{array} \right.\) 

`=> D = (1; +∞)`