a) \(2x+8=0\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
b)\(\left|3x+1\right|=7+x\)
\(\left\{{}\begin{matrix}3x+1=7+x\forall3x+1\ge0\\-3x-1=7+x\forall3x+1\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\forall x\ge-\dfrac{1}{3}\left(n\right)\\x=-2\forall x\le-\dfrac{1}{3}\left(n\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-2\right\}\)
c) \(5x+1< 9-3x\)
\(\Leftrightarrow8x< 8\)
\(\Leftrightarrow x< 1\)
Vậy \(S=\left\{x\in R|x< 1\right\}\)
d) \(\dfrac{-5}{3x+4}>0\)
\(\Leftrightarrow3x+4< 0\) (do \(-5< 0\))
\(\Leftrightarrow x< -\dfrac{3}{4}\)
Vậy \(S=\left\{x\in R|x< -\dfrac{3}{4}\right\}\)
Tick hộ nha 😘
a) \(2x+8=0\)
\(\Leftrightarrow x=-\dfrac{8}{2}\)
\(\Leftrightarrow x=-4\)
Vậy \(S=\left\{-4\right\}\)
b) \(\left|3x+1\right|=7+x\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x-1=7+x\\3x+1=7+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x=8\\2x=6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{-2;3\right\}\)
c) \(5x+1< 9-3x\)
\(\Leftrightarrow8x< 8\)
\(\Leftrightarrow x< 1\)
Vậy \(S=\left\{x/x< 1\right\}\)
d) \(\dfrac{-5}{3x+4}>0\) (*)
ĐKXĐ: \(3x+4\ne0\Leftrightarrow x\ne-\dfrac{4}{3}\)
Ta có \(\dfrac{-5}{3x+4}>0\) mà \(-5< 0\Rightarrow3x+4< 0\) \(\Rightarrow x< -\dfrac{4}{3}\)
Vậy \(S=\left\{x/x< -\dfrac{4}{3}\right\}\)
a) Ta có: 2x+8=0
nên 2x=-8
hay x=-4
d) Ta có: \(\dfrac{-5}{3x+4}>0\)
nên 3x+4<0
hay \(x< -\dfrac{4}{3}\)
