a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

c) \(A=2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\)(vô lý do \(2\sqrt{x}\ge0\forall x\))

Vậy \(S=\varnothing\)

Bài 3:

\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)

\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)

a, ĐKXĐ: \(x^2-3\ge0\Rightarrow x^2\ge3\Rightarrow x\ge\sqrt{3}\)

b, \(\left\{{}\begin{matrix}x-2\ne0\\x-2\ge0\end{matrix}\right.\Rightarrow x-2>0\Rightarrow x>2\)

c, \(\left\{{}\begin{matrix}3-2x\ne0\\\dfrac{1}{3-2x}\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x\ne3\\3-2x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{3}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)

\(\sqrt{x^2-3}\)

ĐKXĐ: x > 1

\(\dfrac{x}{x-2}+\sqrt{x-2}\)

ĐKXĐ: x > 2

\(\sqrt{\dfrac{1}{3-2x^2}}\)

ĐKXĐ: x < 1,224744871 \(\approx\) 1,22

a: ĐKXĐ: x^2-1>=0 

=>x>=1 hoặc x<=-1

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

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

x>=căn 2

=>x^2>=2

=>x^2-1>=1

=>căn x^2-1>=1

=>căn(x^2-1)-1>=0

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

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(P=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)

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

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

b) Để P<1 thì \(2\sqrt{x}< 2\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

ủa bạn ơi, sao cái dấu bằng đầu tiên đã ra như vậy rồi, mình ko hiểu

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge6\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(-1\le x\le1\)

c: ĐKXĐ: \(x\le-2\)

chị giỏi quá

\(ĐK:4-\dfrac{2}{7}x\ge0\Leftrightarrow-\dfrac{2}{7}x\ge-4\Leftrightarrow x\le14\)

x≤14