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

\(< =>1-3x\ge0\)

\(3x\le1\)

\(x\le\frac{1}{3}\)

\(b,-3< 0\)

\(< =>2x-5\ne0;2x-5\le0< =>2x-5< 0\)

\(x< \frac{5}{2}\)

\(c,\sqrt{3x+2}+\sqrt{-2x+3}\)

\(\hept{\begin{cases}3x+2\ge0\\-2x+3\ge0\end{cases}}\)

\(\hept{\begin{cases}x\ge-\frac{2}{3}\\x\le\frac{3}{2}\end{cases}}\)

\(< =>-\frac{2}{3}\le x\le\frac{3}{2}\)

\(d,\frac{x-5}{\sqrt{-4x}}\)

\(\sqrt{-4x}\ge0;\sqrt{-4x}\ne0< =>\sqrt{-4x}>0\)

\(-4x>0\)

\(x< 0\)

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

\(\sqrt{x-2}\ge0;x-3\ne0\)

\(x\ge2;x\ne3\)

\(f,\sqrt{-\left(x-2\right)^2}\)

\(\sqrt{-\left(x-2\right)^2}\ge0\)

\(-\left|x-2\right|\ge0\)

\(-\left|x-2\right|\le0\)

lên chỉ có 1 nghiệm duy nhất là 

\(x-2=0< =>x=2\)

\(g,\sqrt{\frac{-2x^2}{3x+2}}\)

\(-2x^2\le0\)

\(\sqrt{\frac{-2x^2}{3x+2}}\ge0< =>3x+2\le0;3x+2\ne0\)

\(x\le-\frac{2}{3};x\ne-\frac{2}{3}< =>x< -\frac{2}{3}\)

a)\(\sqrt{1-3x}\)có nghĩa \(\Leftrightarrow\sqrt{1-3x}\ge0\)

\(\Leftrightarrow1-3x\ge0\)

\(\Leftrightarrow-3x\ge-1\)

\(\Leftrightarrow x\ge\frac{1}{3}\)

b)\(\sqrt{\frac{-3}{2x-5}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-3}{2x-5}}\ge0\)

\(\Leftrightarrow\frac{-3}{2x-5}\ge0\)

\(\Leftrightarrow2x-5>0\)

\(\Leftrightarrow2x>5\)

\(\Leftrightarrow x>\frac{5}{2}\)

c)\(\sqrt{3x+2}+\sqrt{-2x+3}\)có nghĩa \(\sqrt{3x+2}+\sqrt{-2x+3}\ge0\)

\(\Leftrightarrow3x+2-2x+3\ge0\)

\(\Leftrightarrow x+5\ge0\)

\(\Leftrightarrow x\ge-5\)

d)\(\frac{x-5}{\sqrt{-4x}}\)có nghĩa \(\Leftrightarrow\frac{x-5}{\sqrt{-4x}}\ge0\)

\(\Leftrightarrow\frac{x-5}{\sqrt{-\left(2x\right)^2}}\ge0\)

\(\Leftrightarrow\frac{x-5}{-2x}\ge0\)

\(\Leftrightarrow-2x>0\)

\(\Leftrightarrow x>2\)(Câu này không chắc làm đúng không, chắc sai goi)

f)\(\sqrt{-x^2+4x-4}\)có nghĩa \(\Leftrightarrow\sqrt{-x^2+4x-4}\ge0\)

\(\Leftrightarrow-x^2+4x-4\ge0\)

\(\Leftrightarrow-\left(x-2\right)^2\ge0\)

không có z thỏa mãn

g)\(\sqrt{\frac{-2x^2}{3x+2}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-2x^2}{3x+2}}\ge0\)

\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\)

\(\Leftrightarrow3x+2>0\)

\(\Leftrightarrow3x>-2\)

\(\Leftrightarrow x>\frac{-2}{3}\)

@Cừu

\(a,\frac{1}{\sqrt{5x+15}}\)

Để biểu thức trên có nghĩa :

\(\Rightarrow\sqrt{5x+15}\ge0\)

\(\Rightarrow5\left(x+3\right)\ge0\)

\(\Rightarrow x\ge-3\)

Vậy....

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 :

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

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

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

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

\(\Leftrightarrow A=-2\)