a,

 \(\left(5x+3\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}5x+3=\dfrac{5}{3}\\5x+3=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

b,

\(\left(-\dfrac{1}{2}x+3\right)^3=-\dfrac{1}{125}\\ \Rightarrow-\dfrac{1}{2}x+3=-\dfrac{1}{5}\\ \Rightarrow x=\dfrac{32}{5}\)

c,

Ta có: \(\left|\dfrac{1}{2}x+3\right|\ge0\) với \(\forall x\)

\(\Rightarrow\left|\dfrac{1}{2}x+3\right|-\dfrac{5}{9}\ge-\dfrac{5}{9}\) với \(\forall x\)

  Dấu '' = '' xảy ra khi :

\(\left|\dfrac{1}{2}x+3\right|=0\\ \Rightarrow \dfrac{1}{2}x+3=0\\ \Leftrightarrow x=6\)

    Vậy GTNN của \(\left|\dfrac{1}{2}x+3\right|-\dfrac{5}{9}\) là \(-\dfrac{5}{9}\) khi \(x=6\)

16. B

17. A

18. B

19. D

20. B

\(x^2-8x+16\\ =x^2-4x-4x+16\\=\left(x^2-4x\right)-\left(4x-16\right)\\ =x\left(x-4\right)-4\left(x-4\right)\\ =\left(x-4\right)\left(x-4\right)\\ =\left(x-4\right)^2\)

 \(x^2+4x+4\\ =x^2+2x+2x+4\\ =\left(x^2+2x\right)+\left(2x+4\right)\\ =x\left(x+2\right)+2\left(x+2\right)\\ =\left(x+2\right)\left(x+2\right)\\ =\left(x+2\right)^2\)

\(-6< x< -4\dfrac{2}{5}\) đây ạ !

Mik xin lỗi mik k cố ý bạn thông cảm nhé

À chết bạn sử số 1 chỗ đó thành sô 4 nhé mik bị lỗi ạ

Ta có: \(\left|\dfrac{7}{8}x+\dfrac{5}{6}\right|-\left|\dfrac{1}{2}x+5\right|=0\)

 \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x+\dfrac{5}{6}=0\\\dfrac{1}{2}x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=-\dfrac{5}{6}\\\dfrac{1}{2}x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{20}{21}\\x=-10\end{matrix}\right.\)

    Vậy \(\left[{}\begin{matrix}x=-\dfrac{20}{21}\\x=-10\end{matrix}\right.\)

Ta có :

 \(x-\dfrac{8}{5}< -6\\ \Rightarrow x< -6+\dfrac{8}{5}\\ \Rightarrow x< -\dfrac{22}{5}=-4\dfrac{2}{5}\\ \Rightarrow-6< x< -1\dfrac{2}{5}\\ \Rightarrow x=-5\)

    Vậy...