Bài 8:
a) ĐKXĐ: \(x\ge\dfrac{2019}{3}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}2-5x\ne0\\4-3x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{2}{5}\\x\le\dfrac{4}{3}\end{matrix}\right.\)
c) ĐKXĐ: \(\left\{{}\begin{matrix}2-3x\ge0\\x+5>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x>-5\end{matrix}\right.\Leftrightarrow-5< x\le\dfrac{2}{3}\)
e) ĐKXĐ: \(\dfrac{3x-1}{4-x}\ge0\)
\(\Leftrightarrow\dfrac{3x-1}{x-4}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\x-4< 0\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 4\)
h) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-1\end{matrix}\right.\)
Bài 10:
a) Ta có: \(\left|x-4\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=3\\x-4=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
b) Ta có: \(\sqrt{x^2+3}=\sqrt{4x}\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
c) Ta có: \(x+\sqrt{3x+10}=0\)
\(\Leftrightarrow\sqrt{3x+10}=-x\left(x\le0\right)\)
\(\Leftrightarrow x^2=3x+10\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(loại\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
