1) \(\sqrt{\sqrt{5}-\sqrt{3x}}\) xát định \(\Leftrightarrow\) \(\left\{{}\begin{matrix}3x\ge0\\\sqrt{5}-\sqrt{3x}\ge0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x\ge0\\\sqrt{3x}\le\sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x\ge0\\3x\le5\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x\ge0\\x\le\dfrac{5}{3}\end{matrix}\right.\) \(\Rightarrow\) \(0\le x\le\dfrac{5}{3}\)

2) \(\sqrt{\sqrt{6x}-4x}\) xát định \(\Leftrightarrow\) \(\left\{{}\begin{matrix}6x\ge0\\\sqrt{6x}-4x\ge0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x\ge0\\x\le\dfrac{3}{8}\end{matrix}\right.\) \(\Leftrightarrow\) \(0\le x\le\dfrac{3}{8}\)

3) ta có : \(\left(x-6\right)^6\ge0\forall x\) \(\Rightarrow\) \(\sqrt{\left(x-6\right)^6}\) được xát định \(\forall x\)

4) \(2-4\sqrt{5x+8}\) xát định \(\Leftrightarrow\) \(5x+8\ge0\) \(\Leftrightarrow\) \(5x\ge-8\) \(\Leftrightarrow\) \(x\ge\dfrac{-8}{5}\)

5) \(\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xát định \(\Leftrightarrow\) \(\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}>0\)

mà ta có \(-2\sqrt{6}+\sqrt{23}< 0\) \(\Rightarrow\) để \(\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}>0\)

\(\Leftrightarrow\) \(-x+5< 0\) \(\Leftrightarrow\) \(x>5\) (và \(x\ne5\) )

6) \(\sqrt{\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}}\) xát định \(\Leftrightarrow\) \(\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}>0\)

mà \(2\sqrt{15}-\sqrt{59}>0\) \(\Rightarrow\) để \(\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}>0\)

thì \(x-7>0\) \(\Leftrightarrow\) \(x>7\) (và \(x\ne7\) )

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

mik cần gấp ạ^^

a)đk:`2x-4>=0`

`<=>2x>=4`

`<=>x>=2.`

b)đk:`3/(-2x+1)>=0`

Mà `3>0`

`=>-2x+1>=0`

`<=>1>=2x`

`<=>x<=1/2`

c)`đk:(-3x+5)/(-4)>=0`

`<=>(3x-5)/4>=0`

`<=>3x-5>=0`

`<=>3x>=5`

`<=>x>=5/3`

d)`đk:-5(-2x+6)>=0`

`<=>-2x+6<=0`

`<=>2x-6>=0`

`<=>2x>=6`

`<=>x>=3`

e)`đk:(x^2+2)(x-3)>=0`

Mà `x^2+2>=2>0`

`<=>x-3>=0`

`<=>x>=3`

f)`đk:(x^2+5)/(-x+2)>=0`

Mà `x^2+5>=5>0`

`<=>-x+2>0`

`<=>-x>=-2`

`<=>x<=2`

a, ĐKXĐ : \(2x-4\ge0\)

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

Vậy ..

b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3}{-2x+1}\ge0\\-2x+1\ne0\end{matrix}\right.\)

\(\Leftrightarrow-2x+1>0\)

\(\Leftrightarrow x< \dfrac{1}{2}\)

Vậy ..

c, ĐKXĐ : \(\dfrac{-3x+5}{-4}\ge0\)

\(\Leftrightarrow-3x+5\le0\)

\(\Leftrightarrow x\ge\dfrac{5}{3}\)

Vậy ...

d, ĐKXĐ : \(-5\left(-2x+6\right)\ge0\)

\(\Leftrightarrow-2x+6\le0\)

\(\Leftrightarrow x\ge-\dfrac{6}{-2}=3\)

Vậy ...

e, ĐKXĐ : \(\left(x^2+2\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow x-3\ge0\)

\(\Leftrightarrow x\ge3\)

Vậy ...

f, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{x^2+5}{-x+2}\ge0\\-x+2\ne0\end{matrix}\right.\)

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

\(\Leftrightarrow x< 2\)

Vậy ...

1: ĐKXĐ: 2-3x>=0

=>x<=2/3

2: ĐKXĐ: -3x^2>=0

=>x^2<=0

=>x=0

3: ĐKXĐ: -2023x^3>=0

=>x^3<=0

=>x<=0

4: ĐKXĐ: -2(x-5)>=0

=>x-5<=0

=>x<=5

5: ĐKXĐ: -5/2-2x>=0

=>2-2x<0

=>2x>2

=>x>1

6: ĐKXĐ: (x^2+1)(3-2x)>=0

=>3-2x>=0

=>-2x>=-3

=>x<=3/2

7: ĐKXĐ: (-x^2-1)(3-x)>=0

=>(x^2+1)(x-3)>=0

=>x-3>=0

=>x>=3

\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)

a: ĐKXĐ: x^2-2x<>0 và x^2-1>0

=>(x>1 và x<>2) hoặc x<-1

b: ĐKXĐ: x+1>0 và 5-3x>0

=>x>-1 và 3x<5

=>-1<x<5/3

c: DKXĐ: 5x+3>=0 và 3-x>0

=>x>=-3/5 và x<3

=>-3/5<=x<3

d: ĐKXĐ: 4-x^2>0 và 1+x>=0

=>x^2<4 và x>=-1

=>-2<x<2 và x>=-1

=>-1<=x<2

e: ĐKXĐ: 2-3x<>0 và 1-6x>0

=>x<>2/3 và x<1/6

=>x<1/6