a) ĐKXĐ: \(x\in R\)

b) ĐKXĐ: \(-2\sqrt{2}+2\le x\le2\sqrt{2}+2\)

a) ĐKXĐ: \(x\ge2\)

b) ĐKXĐ: \(\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

c) ĐKXĐ: \(\dfrac{x+3}{5-x}\ge0\)

\(\Leftrightarrow\dfrac{x+3}{x-5}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow-3\le x< 5\)

a) Để \(\sqrt{\dfrac{x}{3}}\) có nghĩa thì \(\dfrac{x}{3}\ge0\Leftrightarrow x\ge0\)

b) Để \(\sqrt{-5x}\) có nghĩa thì \(-5x\ge0\Leftrightarrow x\le0\)

c) Để \(\sqrt{4-x}\) có nghĩa thì \(4-x\ge0\Leftrightarrow x\le4\)

d) Để \(\sqrt{3x+7}\) có nghĩa thì \(3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\)

e) Để \(\sqrt{-3x+4}\) có nghĩa thì \(-3x+4\ge0\Leftrightarrow x\le\dfrac{4}{3}\)

f) Để \(\sqrt{\dfrac{1}{-1+x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\dfrac{1}{-1+x}\ge0\\-1+x\ne0\end{matrix}\right.\)

\(\Leftrightarrow-1+x>0\Leftrightarrow x>1\)

g) Để \(\sqrt{1+x^2}\) có nghĩa thì \(1+x^2\ge0\left(đúng\forall x\right)\)

h) \(\sqrt{\dfrac{5}{x-2}}\) có nghĩ thì \(\left\{{}\begin{matrix}\dfrac{5}{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow x-2>0\Leftrightarrow x>2\)

a. \(x\ge0\)

b. \(x< 0\)

c. \(x\le4\)

d. \(x\ge\dfrac{-7}{3}\)

e. \(x\le\dfrac{4}{3}\)

f. \(x>1\)

g. Mọi x

h. \(x>2\)

a: ĐKXĐ: \(x\ge\dfrac{1}{3}\)

b: ĐKXĐ: \(x< \dfrac{15}{2}\)

c: ĐKXĐ: \(x\le0\)

$a)ĐK:8x+2\ge 0$

$\to 8x \ge -2$

$\to x \ge -\dfrac14$

$b)ĐK:\dfrac{-5}{6-3x} \ge 0(x \ne 2)$

Mà $-5<0$

$\to 6-3x<0$

$\to 6<3x$

$\to x>2$

$*A=x-2\sqrt{x-2}+3(x \ge 2)$

$=x-2-2\sqrt{x-2}+1+4$

$=(\sqrt{x-2}-1)^2+4 \ge 4$

Dấu "=" xảy ra khi $\sqrt{x-2}-1=0 \Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$

a) \(x\ge-\dfrac{1}{4}\)

b) x<2

a: ĐKXĐ: \(x\in R\)

b: ĐKXĐ: x=2

x=2

a) \(\sqrt{x^2-x+1}\)

\(=\sqrt{x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}}\)

\(=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)

Nên bt luôn có nghĩa

b) \(\dfrac{5}{\sqrt{1-\sqrt{x-1}}}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x-1\ge0\\1-\sqrt{x-1}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x-1< 1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\\x< 2\end{matrix}\right.\Leftrightarrow1\le x< 2\)

c) \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) có nghĩa khi:

\(x\ge0\)

d) \(\dfrac{\sqrt{-3x}}{x^2-1}\) có nghĩa khi:

\(\Leftrightarrow\left\{{}\begin{matrix}-3x\ge0\\x^2-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x\ne\pm1\end{matrix}\right.\)

e) \(\dfrac{2}{\sqrt{x}-2}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(\sqrt{\dfrac{3x-2}{x^2-2x+4}}=\sqrt{\dfrac{3x-2}{\left(x-2\right)^2}}\) 

Có nghĩa khi:

\(\left\{{}\begin{matrix}\dfrac{3x-2}{\left(x-2\right)^2}\ge0\\\left(x-2\right)^2\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ne2\end{matrix}\right.\)

____________________

\(\sqrt{\dfrac{2x-3}{2x^2+1}}\)

Có nghĩa khi:

\(\dfrac{2x-3}{2x^2+1}\ge0\)

\(\Leftrightarrow2x-3\ge0\)

\(\Leftrightarrow x\ge\dfrac{3}{2}\)

a: ĐKXĐ: (3x-2)/(x^2-2x+4)>=0

=>3x-2>=0

=>x>=2/3

b: ĐKXĐ: (2x-3)/(2x^2+1)>=0

=>2x-3>=0

=>x>=3/2

giúp mình vs

\(\Leftrightarrow3x-2\ge0\)

hay \(x\ge\dfrac{2}{3}\)