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) Biểu thức xác định `<=> x^2-2x-1>0`

`<=>(x^2-2x+1)-2>0`

`<=>(x-1)^2-(\sqrt2)^2>0`

`<=>(x-1+\sqrt2)(x-1-\sqrt2)>0`

`<=>` \(\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\)

`D=(-∞; 1-\sqrt2) \cup (1+\sqrt2 ; +∞)`

b) Biểu thức xác định `<=> x-\sqrt(2x+1)>0`

`<=> x>\sqrt(2x+1)`

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x+1\ge0\\x^2>2x+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow x>1+\sqrt{2}\)

`D=(1+\sqrt2 ; +∞)`

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

x∈[0, ∞)

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

a) ĐKXĐ: \(\hept{\begin{cases}2x-1\ge0\\2x\ge2\sqrt{2x-1}\end{cases}}\)\(\Leftrightarrow x\ge\frac{1}{2}\)

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

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

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

Nếu \(x\ge1\)thì A=\(\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)\)=2.

Nếu \(\frac{1}{2}\le x< 1\)thì A=\(\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)\)=\(2\sqrt{2x-1}\).

b)A<1 thì \(\frac{1}{2}\le x< 1\)và \(2\sqrt{2x-1}< 1\)\(\Leftrightarrow4\left(2x-1\right)< 1\)\(\Leftrightarrow8x-4< 1\)\(\Leftrightarrow x< \frac{5}{8}\)(tm)

Vậy A<1 thì \(\frac{1}{2}\le x< \frac{5}{8}\).

\(a,\)\(\frac{2}{\sqrt{x^2-x+1}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x^2-x+1\ge0\\x^2-x+1\ne0\end{cases}\Rightarrow x^2-x+1>0}\)

Mà \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)với \(\forall x\)

\(\Rightarrow\)Biểu thức luôn được xác định với mọi x 

\(a,3-\sqrt{1-16x^2}\)

\(=3-\sqrt{-\left(16x^2-1\right)}\)

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

Căn thức xác định khi \(\sqrt{-\left(4x-1\right)\left(4x+1\right)\ge0}\)

\(\Rightarrow\left(4x-1\right)\left(4x+1\right)\le0\)

.....

\(b,\sqrt{2x^2-6}\)

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

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

Để căn thức xác định \(\Rightarrow\sqrt{2\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}\ge0\)

\(\Rightarrow\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\ge0\)

.....

\(b,\sqrt{\frac{2x-1}{x+3}}\)

\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)

Và \(\frac{2x-1}{x+3}\ge0\)

Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)

Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)