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b: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)
\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{0;-1;1;-2\right\}\)
Để \(2x^5+4x^4-7x^3-44⋮2x^2-7\)
\(\Leftrightarrow5⋮2x^2-7\)
\(\Leftrightarrow2x^2-7\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Ta có bảng sau :
\(2x^2-7\) | 1 | -1 | 5 | -5 |
x | \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) | \(\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) | \(\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\) | \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\) |
Vì x là số nguyên \(\Rightarrow x\in\left\{2;-2;1;-1\right\}\)
Vậy \(x\in\left\{2;-2;1;-1\right\}\) thì \(2x^5+4x^4-7x^3-44⋮2x^2-7\)
\(M⋮N\\ \Rightarrow3x^3+4x^2-7x+5⋮x-3\\ \Rightarrow3x^3-9x^2+13x^2-39x+32x-96+101⋮x-3\\ \Rightarrow3x^2\left(x-3\right)+13x\left(x-3\right)+32\left(x-3\right)+101⋮x-3\\ \Rightarrow x-3\inƯ\left(101\right)=\left\{-101;-1;1;101\right\}\\ \Rightarrow x\in\left\{-98;2;4;104\right\}\)
\(\Leftrightarrow8x^3-2x^2-15x+m=\left(4x-3\right)\cdot a\left(x\right)\)
Thay \(x=\dfrac{3}{4}\Leftrightarrow8\cdot\left(\dfrac{3}{4}\right)^3-2\left(\dfrac{3}{4}\right)^2-15\cdot\dfrac{3}{4}+m=0\)
\(\Leftrightarrow8\cdot\dfrac{27}{64}-2\cdot\dfrac{9}{16}-\dfrac{45}{4}+m=0\\ \Leftrightarrow\dfrac{27}{8}-\dfrac{9}{8}-\dfrac{45}{4}+m=0\\ \Leftrightarrow\dfrac{9}{4}-\dfrac{45}{4}+m=0\\ \Leftrightarrow m-9=0\\ \Leftrightarrow m=9\)