a, \(x+1\ge0\Leftrightarrow x\ge-1\)

b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)

d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)

e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)

f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)

g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)

3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)

4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)

5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)

6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)

1: ĐKXĐ: \(a>-2\)

2: ĐKXĐ: \(x\ne2\)

3: ĐKXĐ: \(a\in\varnothing\)

1)
\(-\dfrac{1}{\sqrt{a+2}}\) có nghĩa khi \(\sqrt{a+2}>0\)
=>a+2>0
    a>-2
2)
\(\sqrt{\dfrac{3}{\left(x-2\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(x-2\right)^2}}\) 
mà \(\left(x-2\right)^2>0=>\sqrt{\left(x-2\right)^2}>0vớimọix\)
3)
\(\sqrt{\dfrac{-3}{a^2-4a+4}}=\sqrt{\dfrac{-3}{\left(a-2\right)^2}}cónghĩakhi\left(a-2\right)^2< 0mà\left(a-2\right)^2>0=>biểuthứckocónghĩavớimọia\)
 

a) \(-\dfrac{1}{\sqrt{a+2}}\Rightarrow\sqrt{a+2}>0\Leftrightarrow a>-2\)

b) \(\sqrt{\dfrac{3}{\left(x-2\right)^2}}\Rightarrow x-2\ne0\Leftrightarrow x\ne2\)

c) \(\sqrt{\dfrac{-3}{a^2-4a+4}}\Rightarrow a^2-4a+4< 0\Leftrightarrow a^2-4a< -4\)

d) \(\sqrt{\dfrac{2}{x^2+2x+2}}\Rightarrow x^2+2x+2>0\Leftrightarrow x^2+2x>-2\)

e) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\Rightarrow x^2-4x+5< 0\Leftrightarrow x^2-4x< -5\)

f) \(\sqrt{\dfrac{-4}{x^2-1}}\Rightarrow x^2-1< 0\Rightarrow x^2< 1\Rightarrow x< 1\)

2 câu cuối do lỗi nên mk ko gõ cth được

ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

b) ĐKXĐ: \(-1\le x\le3\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).

d) ĐKXĐ: \(x< \dfrac{3}{5}\).

giúp mình với ahuhuuu

1) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

2) ĐKXĐ: \(x\le\dfrac{3}{2}\)

3) ĐKXĐ: \(x\le-2\)

4) ĐKXĐ: \(x< \dfrac{1}{4}\)

5) ĐKXĐ: \(x\le-\dfrac{5}{3}\)

a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:

\(\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:

\(\left(2x-5\right)\left(2x+5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

1) 

a) \(\sqrt{2x-4}\) có nghĩa khi:

\(2x-4\ge0\)

\(\Leftrightarrow2x\ge4\)

\(\Leftrightarrow x\ge\dfrac{4}{2}\)

\(\Leftrightarrow x\ge2\)

b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi 

\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)

\(\Rightarrow4-x\le0\)

\(\Leftrightarrow x\ge4\)

2) 

a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)

\(A=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(A=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)

\(A=\sqrt{5}+2-\sqrt{5}+2\)

\(A=4\)

\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)

\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=-\left(7-5\right)\)

\(B=-2\)