biểu thứ A có ý nghĩa khi \(\sqrt{4-3x}\ge0\\=>4-3x\ge0\\ =>3x\ge4=>x\ge\dfrac{4}{3}\)

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

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

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

\(=\frac{\sqrt{2\text{x}}+x}{\sqrt{2}+2}.\frac{\sqrt{x}-2}{\sqrt{4\text{x}}}\)

\(=\frac{x\sqrt{2}-2\sqrt{2\text{x}}+x\sqrt{x}-2\text{x}}{2\sqrt{2\text{x}}+4\sqrt{x}}\)

tick cho mình nha

a) \(ĐKXĐ:x\ge0;x\ne3\)

b) \(A=\left(\frac{x-2\sqrt{3x}+3}{x-3}\right)\left(\sqrt{4x}+\sqrt{12}\right)\)

\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-\sqrt{3}\right)^2}{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}\right)\left(2\sqrt{x}+2\sqrt{3}\right)\)

\(\Leftrightarrow A=\left(\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x}+\sqrt{3}}\right).2\left(\sqrt{x}+\sqrt{3}\right)\)

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

\(\Leftrightarrow A=2\sqrt{x}-2\sqrt{3}\)

c) Thay \(x=4-2\sqrt{3}\)vào A, ta có :

\(A=2\sqrt{4-2\sqrt{3}}-2\sqrt{3}\)

\(\Leftrightarrow A=2\sqrt{\left(1-\sqrt{3}\right)^2}-2\sqrt{3}\)

\(\Leftrightarrow A=2\left(\sqrt{3}-1\right)-2\sqrt{3}\)

\(\Leftrightarrow A=2\sqrt{3}-2-2\sqrt{3}\)

\(\Leftrightarrow A=-2\)

Để A có nghĩa thì A≥0

⇒-3/3-x≥0

⇒3-x≤-1

⇒x≤4

ĐKXĐ:

a. 

\(\dfrac{-3}{3-x}\ge0\Rightarrow3-x< 0\Rightarrow x>3\)

b.

\(x+\dfrac{1}{x}\ge0\Rightarrow\dfrac{x^2+1}{x}\ge0\Rightarrow x>0\)

a: ĐKXĐ: \(x>3\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

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

b: Thay x=9 vào A, ta được:

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

c: Ta có: P=AB

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

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{6}{\sqrt{x}+1}\)

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

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)

b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)