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Ta có
x 4 – x 3 y + x 2 y 2 – x y 3 = x 4 + x 2 y 2 – ( x 3 y + x y 3 ) = x 2 ( x 2 + y 2 ) – x y ( x 2 + y 2 ) = ( x 2 + y 2 ) ( x 2 – x y ) = ( x 2 + y 2 ) x ( x – y ) N ê n ( x 4 – x 3 y + x 2 y 2 – x y 3 ) : ( x 2 + y 2 ) = ( x 2 + y 2 ) x ( x – y ) : ( x 2 + y 2 ) = x ( x – y )
Đáp án cần chọn là : B
1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
2, \(5x\left(3y+4x-6\right)\)
3, \(3x\left(2-y\right)\)
4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)
6, \(2xy\left(x+2y-5x^2y\right)\)
7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
11, \(\left(x+y\right)\left(x-1\right)\)
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
e: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
f: \(x^2-2x+7x-14\)
\(=x\left(x-2\right)+7\left(x-2\right)\)
=(x-2)(x+7)
h: \(5x^2-10xy+5y^2-20\)
\(=5\left(x^2-2xy+y^2-4\right)\)
\(=5\left(x-y-2\right)\left(x-y+2\right)\)
a: \(3x^4-6x^3+2x^2=x^2\left(3x^2-6x+2\right)\)
b: \(x^3y+12x^2y+36xy=xy\left(x^2+12x+36\right)=xy\left(x+6\right)^2\)
c: \(x^3y-9xy^3=xy\left(x^2-9y^2\right)=xy\left(x-3y\right)\left(x+3y\right)\)
d: \(x^2y^2-2xy^2+y^2=y^2\left(x-1\right)^2\)
a: =x^3(x-y)+(x-y)
=(x-y)(x^3+1)
=(x-y)(x+1)(x^2-x+1)
b: =(a-1)^2-9b^2
=(a-1-3b)(a-1+3b)
Đặt \(f\left(x\right)=2x^3-3x^2+x+a\)
Ta có: phép chia \(f\left(x\right)\) cho \(x+2\) có dư là \(R=f\left(-2\right)\)
\(\Rightarrow f\left(-2\right)=2.\left(-2\right)^3-3.\left(-2\right)^2+\left(-2\right)+a\)
\(f\left(-2\right)=2.\left(-8\right)-3.4-2+a\)
\(f\left(-2\right)=-16-12-2+a\)
\(f\left(-2\right)=-20+a\)
Để \(f\left(x\right)\) chia hết cho \(x+2\) thì \(R=0\) hay \(f\left(-2\right)=0\)
\(\Rightarrow-20+a=0\Leftrightarrow a=20\)