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Ta có \(P\left(x\right)\equiv\left(ax-3x\right)+\left(b+2\right)[modQ\left(x\right)]\)
Mà: \(P\left(x\right)⋮Q\left(x\right)\)
\(\Rightarrow\left\{{}\begin{matrix}ax-3x=0\\b+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\)
Vậy \(a=3;b=-2\)
Chúc bạn học tốt!
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
\(\left[{}\begin{matrix}A\left(x\right)=x^4-3x^3+ax+b=x^2\left(x^2-3x+4\right)+\left[\left(a-4\right)x+b\right]=B\left(x\right)+f\left(x\right)\left(a\right)\\A\left(x\right)=x^4-3x^3+ax+b=x^2\left(x^2-3x+2\right)+\left[\left(a-2\right)x+b\right]=C\left(x\right)+g\left(x\right)\left(b\right)\end{matrix}\right.\)
a) \(A\left(x\right)⋮B\left(x\right)\Rightarrow f\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}a=4\\b=0\end{matrix}\right.\)
b)\(A\left(x\right)⋮C\left(x\right)\Rightarrow g\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)