Bài 6:
Ta có: \(\left(-\dfrac{1}{9}\right)^{200}=\left(\dfrac{1}{81}\right)^{100}\)
\(\left(-\dfrac{1}{4}\right)^{300}=\left(-\dfrac{1}{64}\right)^{100}\)
mà \(\dfrac{1}{81}>-\dfrac{1}{64}\)
nên \(\left(\dfrac{-1}{9}\right)^{200}>\left(-\dfrac{1}{4}\right)^{300}\)
Bài 5:
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{1}{3}\)
nên \(x=\dfrac{1}{3}+\dfrac{3}{7}\)
hay \(x=\dfrac{16}{21}\)
b) Ta có: \(\dfrac{-5}{3}+\dfrac{1}{4}x=\left(-\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow\dfrac{1}{4}x-\dfrac{5}{3}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{4}x=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{19}{9}\)
hay \(x=\dfrac{19}{9}:\dfrac{1}{4}=\dfrac{76}{9}\)
c) Ta có: \(\left|2x+\dfrac{3}{4}\right|-6=-5\)
\(\Leftrightarrow\left|2x+\dfrac{3}{4}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{4}=-1\\2x+\dfrac{3}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-7}{4}\\2x=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
