Bài II
\(A\left(x\right)⋮B\left(x\right)\Leftrightarrow2x^3+x^2-4x+m=\left(2x-1\right)\cdot a\left(x\right)\)
Thay \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{4}+\dfrac{1}{4}-2+m=0\Leftrightarrow m=\dfrac{3}{2}\)
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Bài I
\(1,\Rightarrow x^2-4x-x^2+6x-9=0\\ \Rightarrow2x=9\Rightarrow x=\dfrac{9}{2}\\ 2,\Rightarrow x^2-3x-10=0\\ \Rightarrow x^2+2x-5x-10=0\\ \Rightarrow\left(x+2\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
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