## Giúp em với ạ, hứa vote ctlhn

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Giúp em với ạ, hứa vote ctlhn

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4 years 2020-11-04T13:09:19+00:00 2 Answers 73 views 0

1. $\begin{array}{l}a)\,\,y = \dfrac{\sqrt{1 + x}}{x – 1}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}1 + x \geq 0\\x – 1 \ne 0\end{cases}\Leftrightarrow \begin{cases}x \geq – 1\\x \ne 1\end{cases}\\ \Rightarrow TXĐ: D = [-1;+\infty)\backslash \left\{1\right\}\\ b)\,\,y = \dfrac{-x-1}{\sqrt{x-1} – 2\sqrt{2-x}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases} x – 1 \geq 0\\2 – x \geq 0\\\sqrt{x – 1} \ne 2\sqrt{2 – x}\end{cases}\Leftrightarrow \begin{cases}1 \leq x \leq 2\\x \ne \dfrac{9}{5}\end{cases}\\ \Rightarrow TXD: D = [1;2]\backslash\left\{\dfrac{9}{5}\right\}\\ f)\,\,y = \dfrac{1 + \sqrt{2x}}{\sqrt{4 – x}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases} x \geq 0\\4 – x > 0\end{cases}\Leftrightarrow 0 \leq x < 4\\ \Rightarrow TXĐ: D = [0;4)\\ g)\,\,y = \dfrac{1 + \sqrt{x^2 + 2}}{2\sqrt{3 – x}}\\ \text{y xác định}\,\Leftrightarrow 3 – x > 0 \Leftrightarrow x < 3\\ \Rightarrow TXĐ: D = (-\infty;3)\\ l)\,\,y = \dfrac{x-4}{\sqrt{x-1} – \sqrt3}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}x – 1 \geq 0\\\sqrt{x – 1} \ne \sqrt3\end{cases} \Leftrightarrow \begin{cases}x \geq 1\\x \ne 4\end{cases}\\ \Rightarrow TXĐ: D = [1;+\infty)\backslash\left\{4\right\}\\ m)\,\,y = \sqrt{5 – x} + \dfrac{2}{\sqrt{x+1}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}5 – x \geq 0\\x + 1 > 0\end{cases}\Leftrightarrow -1 < x \leq 5\\ \Rightarrow TXĐ: D = (-1;5]\end{array}$

2. a)

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

<=> $$\left\{ \begin{array}{l}x ≥ -1\\x > 1\end{array} \right.$$

=> D = (1; +∞)

b)

$$\left\{ \begin{array}{l}2x ≥ 0\\4 – x > 0\end{array} \right.$$

<=> $$\left\{ \begin{array}{l}x ≥ 0\\x < 4\end{array} \right.$$

=> D = [0; 4)

c)

sqrt{x – 1} – sqrt{3} ne 0

<=> x – 1 ne 3

<=> x ne 4

=> D = [1; +∞) \{4}

d)

$$\left\{ \begin{array}{l}x – 1 > 0\\2 – x > 0\\x – 1 \ne 4(2 – x)\end{array} \right.$$

<=> $$\left\{ \begin{array}{l}x > 1\\x < 2\\x \ne \dfrac{9}{5}\end{array} \right.$$

<=> D = (1; 2) \ {9/5}

e)

3 – x > 0

<=> x < 3

=> D = (3; +∞)

f)

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

<=> $$\left\{ \begin{array}{l}x ≤ 5\\x > -1\end{array} \right.$$

=> D = (-1; 5]