ĐKXĐ : x + 2 \(\ne\)0 => x \(\ne\)-2

Vậy tập xác định : D = R/ {-2}

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

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

\(3x\le1\)

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

\(b,-3< 0\)

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

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

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

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

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

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

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

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

\(-4x>0\)

\(x< 0\)

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

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

\(x\ge2;x\ne3\)

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

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

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

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

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

\(\sqrt{5-3x}\) \(\ge0\) \(\Leftrightarrow5-3x\ge0\Leftrightarrow x\le\frac{5}{3}\)

\(\frac{1}{3x-1}\ge0\Leftrightarrow3x-1\ge0\Leftrightarrow x\ge\frac{1}{3}\)

\(\sqrt{\left(1-x\right)\left(1+x\right)}\ge0\)

\(\Leftrightarrow-1\le x\le1\)

Cau 1 

Dk:5-3x > =0<=> x >=3/5

D = [3/5 ;+oo]

Lời giải:

a) ĐKXĐ: $5-4x\geq 0\Leftrightarrow x\leq \frac{5}{4}$

b) ĐKXĐ: \(\left\{\begin{matrix} 3x-4\neq 0\\ \frac{-5}{3x-4}\geq 0\end{matrix}\right.\Leftrightarrow 3x-4< 0\Leftrightarrow x< \frac{4}{3}\)

c) ĐKXĐ: $x^2+7\geq 0\Leftrightarrow x\in\mathbb{R}$

d)

ĐKXĐ: \(x^2-4x+4\geq 0\Leftrightarrow (x-2)^2\geq 0\Leftrightarrow x\in\mathbb{R}\)

n)

\(\left\{\begin{matrix} x+1\neq 0\\ \frac{3x-5}{x+1}\geq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} 3x-5\geq 0\\ x+1>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x-5\leq 0\\ x+1< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x\geq \frac{5}{3}\\ x< -1\end{matrix}\right.\)

m)

ĐKXĐ: \(\left\{\begin{matrix} 3x-1\neq 0\\ \frac{x^2}{3x-1}\geq 0\end{matrix}\right.\Leftrightarrow 3x-1>0\Leftrightarrow x>\frac{1}{3}\)

g)

ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 5-2x>0\end{matrix}\right.\Leftrightarrow 1\leq x< \frac{5}{2}\)

Em chỉ biết làm câu a thôi :

Mẫu của phân thức A dương mà tử âm nên A_min khi mẫu nhỏ nhất .Ta có :

\(\frac{x^2}{8}-2x+17=\left(\frac{x}{2\sqrt{2}}\right)^2-2.\frac{x}{2\sqrt{2}}.2\sqrt{2}+\left(2\sqrt{2}\right)^2+9\)

\(=\left(\frac{x}{2\sqrt{2}}-2\sqrt{2}\right)^2+9\ge9\Rightarrow\sqrt{\frac{x^2}{8}-2x+17}\ge\sqrt{9}=3\Rightarrow A_{min}=\frac{-3}{3}=-1\)khi :

\(\left(\frac{x}{2\sqrt{2}}-2\sqrt{2}\right)^2=0\Rightarrow\frac{x}{2\sqrt{2}}=2\sqrt{2}\Rightarrow x=8\)