1:

a: ĐKXĐ: 1-x>=0

=>x<=1

b: ĐKXĐ: 2/x>=0

=>x>0

c: ĐKXĐ: 4/x+1>=0

=>x+1>0

=>x>-1

d: ĐKXĐ: x^2+2>=0

=>x thuộc R

Câu 2:

a: \(=\left|-\sqrt{2-1}\right|=\sqrt{1}=1\)

b: \(=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

a)x≥-2

b)x≥1/6

c)x≥0,x≥3/2

\(x^2-25\ge0\Leftrightarrow\left(x-5\right)\left(x+5\right)\ge0\Rightarrow\left[{}\begin{matrix}x\le-5\\x\ge5\end{matrix}\right.\)

\(\frac{x+1}{x-2}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x>2\end{matrix}\right.\)

a: ĐKXĐ: x-2>=0 và 6-2x>=0

=>2<=x<=3

b: DKXĐ: x+2>=0

=>x>=-2

ĐKXĐ: \(x^2+1\ge1\Leftrightarrow x\ge0\)

\(\sqrt{x}=\sqrt{2}-1\)

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

hay \(x=3-2\sqrt{2}\)

ĐK: `x>=0`

`\sqrt2-1 = \sqrtx`

`<=>(\sqrt2-1)^2=x` 

`<=>x=3-2\sqrt2` (TM)

Vậy `S={3-2\sqrt2}`.

Có : `\sqrt{x}=\sqrt{2}-1`

`<=>\sqrt{x}=\sqrt{(\sqrt{2}-1)^2}`

`<=>\sqrt{x}=\sqrt{3-2\sqrt{2}}`

`<=>x=3-2\sqrt{2}`

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

2) ĐKXĐ: \(\dfrac{x-6}{x-2}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2< 0\\x-6\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x\ge6\end{matrix}\right.\)

3) ĐKXĐ: \(\dfrac{2x-4}{5-x}\ge0\)

\(\Leftrightarrow\dfrac{x-2}{x-5}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow2\le x< 5\)

GIÚP VỚI Ạ