\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne3;x\ne-3;x\ne9;x\ne4\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\\ =\dfrac{3}{\sqrt{x}-2}\)

Tick hộ nha 😘

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;1\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{x-4\sqrt{x}+3-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2x-4\sqrt{x}+5-2x+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)

b: Để A>2 thì A-2>0

=>\(\dfrac{1-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)

=>\(\dfrac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)

=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}< 0\)

TH1: \(\left\{{}\begin{matrix}2\sqrt{x}-5>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>\dfrac{5}{2}\\\sqrt{x}< 2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}2\sqrt{x}-5< 0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< \dfrac{5}{2}\\\sqrt{x}>2\end{matrix}\right.\)

=>\(2< \sqrt{x}< \dfrac{5}{2}\)

=>4<x<25/4

c: Để A là số nguyên thì \(1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1\right\}\)

=>\(x\in\left\{1;9\right\}\)

kết hợp ĐKXĐ, ta được: x=9

ĐK:\(\left\{{}\begin{matrix}x+3\ge0\\1-x\ge0\end{matrix}\right.\)\(\Leftrightarrow-3\le x\le1\)

Để biểu thức có nghĩa thì \(\left\{{}\begin{matrix}x+3>0\\1-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x< 1\end{matrix}\right.\Leftrightarrow-3< x< 1\)

Biểu thức trên có nghĩa khi \(\left\{{}\begin{matrix}x+3\ge0\\1-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le1\end{matrix}\right.\)

a: ĐKXĐ: x=0; x<>1

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

\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

a) \(x\ge0\)

b) \(x\le0\)

c) \(x\le4\)

d) \(\sqrt{x^2+1}>0\forall x\) => \(x\in R\)

a)đẻ \(\sqrt{\dfrac{x}{3}}\) có nghĩa thì

\(\dfrac{x}{3}\ge0\\ \Leftrightarrow x\ge0\)

b) để \(\sqrt{-5x}\) có nghĩa thì 

\(-5x\ge0\\ \Leftrightarrow x\le0\)

c) để \(\sqrt{4-x}\) có nghĩa thì 

\(4-x\ge0\\ \Leftrightarrow x\le4\)

d) để \(\sqrt{1+x^2}\) có nghĩa thì

\(1+x^2\ge0\forall x\in R\)

1: ĐKXĐ: 3x^2-x+2>=0

=>x thuộc R

2: ĐKXĐ: x>=0 và căn x-1<>0 và 2-căn x<>0 và 2x+1>0 và x<>0

=>x>0 và x<>1 và x<>4