20 tháng 11 2023 lúc 20:07
TXĐ: D=R\{1/5}
\(y=\dfrac{4x+3}{5x-1}\)
\(y'=\dfrac{\left(4x+3\right)'\left(5x-1\right)-\left(4x+3\right)\left(5x-1\right)'}{\left(5x-1\right)^2}\)
\(\Leftrightarrow y'=\dfrac{4\left(5x-1\right)-5\left(4x+3\right)}{\left(5x-1\right)^2}\)
=>\(y'=\dfrac{20x-4-20x-15}{\left(5x-1\right)^2}=-\dfrac{19}{\left(5x-1\right)^2}\)<0 với mọi x thuộc TXĐ
=>Chọn C
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