Bài 1/
a/ ĐKXĐ: \(\dfrac{-a}{3}\ge 0↔-a\ge 0↔a\le 0\)
b/ ĐKXĐ: \(\dfrac{1}{a}\ge 0\)
mà \(1>0→a>0\)
c/ ĐKXĐ: \(\dfrac{(1-a)^3}{a^2}\ge 0\\→\begin{cases}1-a\ge 0\\a>0\end{cases}\\↔\begin{cases}a\le 1\\a>0\end{cases}\\→0<a\le 1\)
d/ ĐKXĐ: \(\dfrac{a^2+1}{1-2a}\ge 0\)
mà \(a^2+1>0→1-2a>0\\↔2a<1\\↔a<\dfrac{1}{2}\)
e/ ĐKXĐ: \(a^2-1\ge 0\\↔(a-1)(a+1)\ge 0\\↔\left[\begin{array}{1}\begin{cases}a-1\ge 0\\a+1\ge 0\end{cases}\\\begin{cases}a-1\le 0\\a+1\le 0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}a\ge 1\\a\ge -1\end{cases}\\\begin{cases}a\le 1\\a\le -1\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}a\ge 1\\a\le -1\end{array}\right.\)
f/ ĐKXĐ: \(\dfrac{2a-1}{2-a}\ge 0\\↔\left[\begin{array}{1}\begin{cases}2a-1\ge 0\\2-a>0\end{cases}\\\begin{cases}2a-1\le 0\\2-a<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}2a\ge 1\\a<2\end{cases}\\\begin{cases}2a\le 1\\a>2\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}a\ge \dfrac{1}{2}\\a<2\end{cases}\\\begin{cases}a\le \dfrac{1}{2}\\a>2\end{cases}\end{array}\right.\\↔\dfrac{1}{2}\le a<2\)
Bài 2/
a) ĐKXĐ: \(x\in\Bbb R\)
\(\sqrt{(x-3)^2}=3-x\\↔|x-3|=3-x\)
\(|x-3|=\begin{cases}x-3\,\,nếu\,\,x-3\ge 0\,\,hay\,\,x\ge 3\\3-x\,\,nếu\,\,x-3<0\,\,hay\,\,x<3\end{cases}\)
TH1: \(x\ge 3→x-3=3-x\\↔2x=6\\↔x=3(TM)\)
TH2: \(x<3→3-x=3-x\\↔0=0(lđ)\)
Suy ra: \(S=\{x|x\le 3\}\)\)
Vậy \(S=\{x|x\le 3\}\)
b) ĐKXĐ: \(25-20x+4x^2\ge 0\\↔(2x)^2-2.2x.5+5^2\ge 0\\↔(2x-5)^2\ge 0\\→x∈\Bbb R\)
\(\sqrt{25-20x+4x^2}+2x=5\\↔\sqrt{(2x-5)^2}+2x=5\\↔|2x-5|+2x=5\)
\(|2x-5|=\begin{cases}2x-5\,\,nếu\,\,2x-5\ge 0\,\,hay\,\,x\ge \dfrac{5}{2}\\5-2x\,\,nếu\,\,2x-5<0\,\,hay\,\,x<\dfrac{5}{2}\end{cases}\)
TH1: \(x\ge \dfrac{5}{2}→2x-5+2x=5\\↔4x=10\\↔x=\dfrac{5}{2}(TM)\)
TH2: \(x<\dfrac{5}{2}→5-2x+2x=5\\↔5=5(lđ)\)
Suy ra: \(S=\{x|x\le \dfrac{5}{2}\}\)
Vậy \(S=\{x|x\le \dfrac{5}{2}\}\)
