giả sử A>B
\(\Leftrightarrow\dfrac{a}{2a+1}>\dfrac{3x+1}{6x+3}\)
\(\Leftrightarrow\dfrac{3a}{3\left(2x+1\right)}>\dfrac{3x+1}{6x+3}\)
\(\Leftrightarrow\dfrac{3a}{6a+3}>\dfrac{3x+1}{6x+3}\)
\(\Leftrightarrow3a>3a+1\)
\(\Leftrightarrow0>1\) ( vô lí )
vậy \(\Rightarrow B>A\)
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giả sử A>B
⇔a2a+1>3x+16x+3⇔a2a+1>3x+16x+3
⇔3a3(2x+1)>3x+16x+3⇔3a3(2x+1)>3x+16x+3
⇔3a6a+3>3x+16x+3⇔3a6a+3>3x+16x+3
⇔3a>3a+1⇔3a>3a+1
⇔0>1⇔0>1 ( vô lí )
vậy ⇒B>A
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