a: ĐKXĐ: x>=-2/3
\(\dfrac{1}{5}+\sqrt{3x+2}=1\)
=>\(\sqrt{3x+2}=1-\dfrac{1}{5}=\dfrac{4}{5}\)
=>\(3x+2=\dfrac{16}{25}\)
=>\(3x=\dfrac{16}{25}-2=-\dfrac{34}{25}\)
=>\(x=-\dfrac{34}{75}\left(nhận\right)\)
b: ĐKXĐ: x>=-1
\(3\sqrt{x+1}-\dfrac{1}{4}=-5\)
=>\(3\sqrt{x+1}=-5+\dfrac{1}{4}=-\dfrac{19}{4}\)
=>\(\sqrt{x+1}=-\dfrac{19}{12}\left(loại\right)\)
Vậy: \(x\in\varnothing\)
c: \(\dfrac{2}{9}-\sqrt{\left(x+1\right)^2}=-\dfrac{23}{9}\)
=>\(\sqrt{\left(x+1\right)^2}=\dfrac{2}{9}+\dfrac{23}{9}=\dfrac{25}{9}\)
=>\(\left|x+1\right|=\dfrac{25}{9}\)
=>\(\left[{}\begin{matrix}x+1=\dfrac{25}{9}\\x+1=-\dfrac{25}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{25}{9}-1=\dfrac{16}{9}\\x=-\dfrac{25}{9}-1=-\dfrac{34}{9}\end{matrix}\right.\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}2x+1>=0\\x>=0\end{matrix}\right.\)
=>\(x>=0\)
\(\left(\sqrt{2x+1}-4\right)\left(5+\sqrt{x}\right)=0\)
mà \(5+\sqrt{x}>=5>0\forall x\)
nên \(\sqrt{2x+1}-4=0\)
=>\(\sqrt{2x+1}=4\)
=>2x+1=16
=>2x=15
=>\(x=\dfrac{15}{2}\left(nhận\right)\)