Câu 8: \(\left(2x+\dfrac{1}{5}\right)\left(-\dfrac{3}{5}x+\dfrac{4}{7}\right)=0\)
=>\(\left[{}\begin{matrix}2x+\dfrac{1}{5}=0\\-\dfrac{3}{5}x+\dfrac{4}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{5}\\-\dfrac{3}{5}x=-\dfrac{4}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=\dfrac{4}{7}:\dfrac{3}{5}=\dfrac{4}{7}\cdot\dfrac{5}{3}=\dfrac{20}{21}\end{matrix}\right.\)
=>Chọn B
Câu 9: \(C=\dfrac{2^5\cdot5^5-10^6}{3\cdot5^5}=\dfrac{10^5-10^6}{3\cdot5^5}\)
\(=\dfrac{10^5\left(1-10\right)}{3\cdot5^5}=-\dfrac{9}{3}\cdot2^5=-3\cdot32=-96\)
=>Chọn A
Câu 10: D
Câu 11:
\(\left(\dfrac{25}{49}\right)^4=\left[\left(\dfrac{5}{7}\right)^2\right]^4=\left(\dfrac{5}{7}\right)^{2\cdot4}=\left(\dfrac{5}{7}\right)^8\)
=>Chọn D
