\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

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

a: ĐKXĐ: \(-\dfrac{\sqrt{6}}{2}\le x\le\dfrac{\sqrt{6}}{2}\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

c: ĐKXĐ: \(-\sqrt{5}< x< \sqrt{5}\)

d: ĐKXĐ: \(x\le\sqrt[3]{-5}\)

a) Biểu thức xác định `<=> (x+2)(x-1) >=0 <=>` \(\left\{{}\begin{matrix}x\ge1\\x\le-2\end{matrix}\right.\)

b) Biểu thức xác định `<=> (x-3)/(2x-1) >= 0 <=>` \(\left\{{}\begin{matrix}x\ge0\\x< \dfrac{1}{2}\end{matrix}\right.\)

c) Biểu thức xác định `<=> -x^2+2x-1 >= 0 <=> -(x-1)^2 >= 0 <=> x =1`

a) Ko dùng ngoặc nhọn vì không có số nào thỏa mãn \(-2\ge x\ge1\)

b) Biểu thức xác định \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x< \dfrac{1}{2}\end{matrix}\right.\)

\(\sqrt{-x^2+5x-4}+\dfrac{1}{2x-7}\)

Được xác định khi:

\(\left\{{}\begin{matrix}-x^2+5x-4\ge0\\2x-7\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\left(x-4\right)\left(x-1\right)\ge0\\2x\ne7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\left(x-4\right)\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}-\left(x-4\right)< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}-x\ge-4\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}-x< -4\\x< 1\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\le4\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\end{matrix}\right.\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le4\\x\ne\dfrac{7}{2}\end{matrix}\right.\)

\(\sqrt{\dfrac{4}{2x+3}}\) xác định khi \(\dfrac{4}{2x+3}\ge0\Rightarrow2x+3>0\Rightarrow x>-\dfrac{3}{2}\)

\(\sqrt{\dfrac{2x-1}{2-x}}\) xác định khi \(\dfrac{2x-1}{2-x}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2-x< 0\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow\dfrac{1}{2}\le x< 2\)

a/ ĐKXĐ : \(-2x+3\ge0\)

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

b/ ĐKXĐ : \(3x+4\ge0\)

\(\Leftrightarrow x\ge-\dfrac{4}{3}\)

c/ Căn thức \(\sqrt{1+x^2}\) luôn được xác định với mọi x

d/ ĐKXĐ : \(-\dfrac{3}{3x+5}\ge0\)

\(\Leftrightarrow3x+5< 0\)

\(\Leftrightarrow x< -\dfrac{5}{3}\)

e/ ĐKXĐ : \(\dfrac{2}{x}\ge0\Leftrightarrow x>0\)

P.s : không chắc lắm á!

a) ĐK: x ≥ 2

\(\sqrt{3x-6}=3\)

\(\Leftrightarrow3x-6=9\)

<=> 3x = 15

<=> x = 5

Vậy:....

b) ĐK: 5x - 16 ≥ 0

<=> 5x ≥ 16

<=> x ≥ 16/5

\(\sqrt{5x-16}=2\)

<=> 5x - 16 = 4

<=> 5x = 20

<=> x = 4

c) ĐK: \(x^2-4x+3\ne0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)

bình phương hai vế ta được:

a)điều kiện của x:x≥2

3x-6=9 <=> x=5(nhận)

b)ĐK: x≥16/5

5x-16=4 <=>x=4(nhận)

c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)

ĐKXĐ: x≠3 ;x≠1