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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 \(2x^5+4x^4-7x^3-44⋮2x^2-7\)
a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)
\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1\right\}\)
b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)
\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)
\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)
\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)
\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)
\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)
hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)
a) Ta thực hiện phép chia \(3x^3+13x^2-7x+5\) cho \(3x-2\). Khi đó ta có:
\(A=\frac{3x^3+13x^2-7x+5}{3x-2}=3x^2+5x+1+\frac{7}{3x-2}\)
Nếu x nguyên thì \(3x^2+5x+1\in\text{Z}\) nên để A nguyên thì \(\frac{7}{3x-2}\in Z\)
\(\Rightarrow3x-2\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{1;3\right\}\)
b) Ta có: \(B=\frac{2x^5+4x^4-7x^3-44}{2x^2-7}=\left(x^3+2x^2+7\right)+\frac{5}{2x^2-7}\)
Để B nguyên thì \(\frac{5}{2x^2-7}\in Z\Rightarrow2x^2-7\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-1;1;2;-2\right\}\)
b: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)
\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{0;-1;1;-2\right\}\)
a)2x(2x+7)=4(2x+7)
2x(2x+7)-4(2x+7)=0
(2x+7)(2x-4)=0
\(\Rightarrow\orbr{\begin{cases}2x+7=0\\2x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=2\end{cases}}\)
b)Ta có:x3-4x2+ax=x3-3x2-x2+ax
=x2(x-3)-x(x-a)
Để x3-4x2+ax chia hết cho x-3 thì a=3
a: \(2x^5+4x^4-7x^3-44⋮2x^2-7\)
\(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49+5⋮2x^2-7\)
\(\Leftrightarrow2x^2-7\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{2;-2;1;-1\right\}\)
b: \(2x^2+3x+3⋮2x-1\)
\(\Leftrightarrow2x^2-x+4x-2+5⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{1;0;3;-2\right\}\)