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Bài 1 :
Gọi f( x ) = 2n2 + n - 7
g( x ) = n - 2
Cho g( x ) = 0
\(\Leftrightarrow\)n - 2 = 0
\(\Rightarrow\)n = 2
\(\Leftrightarrow\)f( 2 ) = 2 . 22 + 2 - 7
\(\Rightarrow\)f( 2 ) = 3
Để f( x ) \(⋮\)g( x )
\(\Rightarrow\)n - 2 \(\in\)Ư( 3 ) = { \(\pm\)1 ; \(\pm\)3 }
Ta lập bảng :
| n - 2 | 1 | - 1 | 3 | - 3 |
| n | 3 | 1 | 5 | - 1 |
Vậy : n \(\in\){ - 1 ; 1 ; 3 ; 5 }
a: \(\dfrac{2x^3-x^2+ax+b}{x^2-1}\)
\(=\dfrac{2x^3-2x-x^2+1+\left(a+2\right)x+b-1}{x^2-1}\)
\(=2x-1+\dfrac{\left(a+2\right)x+b-1}{x^2-1}\)
Để đây là phép chia hết thì a+2=0 và b-1=0
=>a=-2; b=1
b: \(\Leftrightarrow x^4-1+ax^2-a+bx+a⋮x^2-1\)
=>bx+a=0
=>a=b=0
a) Đặt \(f\left(x\right)=x^3+ax+b\)
Vì \(f\left(x\right)⋮x^2+x-2\)
\(\Rightarrow f\left(x\right)=\left(x^2+x-2\right)q\left(x\right)\)
\(=\left(x^2-x+2x-2\right)q\left(x\right)\)
\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]q\left(x\right)\)
\(=\left(x-1\right)\left(x+2\right)q\left(x\right)\)
\(\Rightarrow f\left(1\right)=\left(1-1\right)\left(1+2\right)q\left(1\right)\)
\(\Rightarrow f\left(1\right)=0\left(1\right)\)
\(f\left(-2\right)=\left(-2-1\right)\left(-2+2\right)q\left(-2\right)\)
\(\Rightarrow f\left(-2\right)=0\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+a+b=0\\-8-2a+b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-1\\-2a+b=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=2\end{matrix}\right.\)
Vậy a=-3 và b=2 thì \(\left(x^3+ax+b\right)⋮\left(x^2+x-2\right)\)
g ) \(4x^2\left(x-2y\right)-\left(4x+1\right)\left(2y-x\right)\)
\(=4x^2\left(x-2y\right)+\left(4x+1\right)\left(x-2y\right)\)
\(=\left(4x^2+4x+1\right)\left(x-2y\right)\)
\(=\left(2x+1\right)^2\left(x-2y\right)\)
h ) \(x^2-ax^2-y+ay+cx^2-cy\)
\(=x^2\left(1-a+c\right)-y\left(1-a+c\right)\)
\(=\left(x^2-y\right)\left(1-a+c\right)\)
\(G=\left[\dfrac{1}{2}ax-2\left(ax+3\right)\right]-\left(ax+3\right)\)
\(=\dfrac{1}{2}ax-2ax-6-ax-3\)
\(=\left(\dfrac{1}{2}ax-2ax-ax\right)+\left(-6-3\right)\)
\(=-\dfrac{5}{2}ax-9\)
\(H=ax-2-\left[-\left(ax-1\right)+3\right]-4\)
\(=ax-2-\left(-ax+4\right)-4\)
\(=ax-2+ax-4-4\)
\(=2ax-10\)
\(G+H=-\dfrac{5}{2}ax-9+2ax-10=-\dfrac{1}{2}ax-19\)
\(G-H=\left(-\dfrac{5}{2}ax-9\right)-\left(2ax-10\right)\)
\(=-\dfrac{5}{2}ax-9-2ax+10\)
\(=-\dfrac{9}{2}ax+1\)