a) Đkxđ: $x^2-2x-3 \ne 0 ⇔$ phương trình $x^2-2x-3=0$ vô nghiệm ⇔ $x \ne 3$ và $x \ne -1$ Vậy TXĐ: D=R\{3;-1} b) Đkxđ: $3x+5≥0 ⇔ x≥\frac{-5}{3}$ $x+5 \ne 0 ⇔ x \ne -5$ Vậy TXĐ: D=[$\frac{-5}{3}$;+∞) c) Đkxđ: $3-2x≥0 ⇔ x≤1,5$ $x+1≥0 ⇔ x≥-1$ Vậy TXĐ: D=[-1;1,5] Log in to Reply
`a) x^2 – 2x – 3 ne 0` `<=>` \(\left[ \begin{array}{l}x \ne 3\\x \ne -1\end{array} \right.\) `=> D = RR \\ {-1; 3}` `b)` \(\left[ \begin{array}{l}x ≥ -\dfrac{5}{3}\\x < -5\end{array} \right.\) `=> D = (-∞; -5) ∪ [-5/3; +∞)` `c)` \(\left[ \begin{array}{l}3 – 2x ≥ 0\\x + 1 > 0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x ≤ \dfrac{3}{2}\\x > -1\end{array} \right.\) `=> D = (-1; 3/2]` Log in to Reply
a) Đkxđ: $x^2-2x-3 \ne 0 ⇔$ phương trình $x^2-2x-3=0$ vô nghiệm
⇔ $x \ne 3$ và $x \ne -1$
Vậy TXĐ: D=R\{3;-1}
b) Đkxđ:
$3x+5≥0 ⇔ x≥\frac{-5}{3}$
$x+5 \ne 0 ⇔ x \ne -5$
Vậy TXĐ: D=[$\frac{-5}{3}$;+∞)
c) Đkxđ:
$3-2x≥0 ⇔ x≤1,5$
$x+1≥0 ⇔ x≥-1$
Vậy TXĐ: D=[-1;1,5]
`a) x^2 – 2x – 3 ne 0`
`<=>` \(\left[ \begin{array}{l}x \ne 3\\x \ne -1\end{array} \right.\)
`=> D = RR \\ {-1; 3}`
`b)`
\(\left[ \begin{array}{l}x ≥ -\dfrac{5}{3}\\x < -5\end{array} \right.\)
`=> D = (-∞; -5) ∪ [-5/3; +∞)`
`c)`
\(\left[ \begin{array}{l}3 – 2x ≥ 0\\x + 1 > 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x ≤ \dfrac{3}{2}\\x > -1\end{array} \right.\)
`=> D = (-1; 3/2]`