a: \(\dfrac{x^2-2x+1}{2x^2-x-1}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
\(2x^2-7x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
hay x=3
\(\Leftrightarrow A=\dfrac{3-1}{2\cdot3+1}=\dfrac{2}{7}\)
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