phân tích các đa thức sau thành nhân tử:
a,3x2-6xy+3y2
b,xy-9x+y-9
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\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)
\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)
\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)
\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)
Lời giải:
a.
$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$
b.
$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$
$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$
$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$
$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$
c.
$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$
$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$
$=(x-1)(x^2+4x+7)$
a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)
\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)
b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)
\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)
\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)
c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)
\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)
a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)
\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)
\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)
b: \(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
b: Ta có: \(xy-3x-2y+6\)
\(=x\left(y-3\right)-2\left(y-3\right)\)
\(=\left(y-3\right)\left(x-2\right)\)
\(a,=y\left(y-2\right)\\ b,=3x\left(x^2-2x+1\right)=3x\left(x-1\right)^2\\ c,=\left(y-1\right)\left(27x^2+9x^3\right)=9x^2\left(x+3\right)\left(y-1\right)\\ d,=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\\ e,=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\\ f,=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\\ g,=\left(2-x\right)\left(x+1\right)\\ h,=\left(x-1\right)\left(3x-6\right)=3\left(x-1\right)\left(x-2\right)\)
a: =y(y-2)
b: \(=3x^2\left(x^2-2x+1\right)=3x^2\left(x-1\right)^2\)
d: \(=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\)
a) \(4x^3y^2-8x^2y+12xy^2=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x-2y\right)^2-49=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-6x+9\right)-25=\left(x-3\right)^2-25=\left(x-3-5\right)\left(x-3+5\right)=\left(x-8\right)\left(x+2\right)\)
a) 4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)
b) 3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)
c) x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)
d) x2−6x−16=(x2−6x+9)−25=(x−3)2−25=(x−3−5)(x−3+5)=(x−8)(x+2)
a) \(4x^3y^2-8x^2y+12xy^2=4xy.x^2y-4xy.2x+4xy.3y=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=\left(3x^2-6xy\right)-\left(5x-10y\right)=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x^2-4xy+4y^2\right)-49=\left(x-2y\right)^2-7^2=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-8x\right)+\left(2x-16\right)=x\left(x-8\right)+2\left(x-8\right)=\left(x-8\right)\left(x+2\right)\)
a,3x2-6xy+3y2
= 3(x2- 2xy+ y2)
= 3(x- y)2
b,xy-9x+y-9
= (xy+ y)- (9x+ 9)
= y(x+ 1)- 9(x+ 1)
= (x+1)(y- 9)
Chúc bạn học tốt
a,\(3x^2-6xy+3y^2\)
=\(3\left(x^2-2xy+y^2\right)\)
=\(3\left(x-y\right)^2\)
b,xy-9x+y-9
=\(\left(xy+y\right)-\left(9x+9\right)\)
=\(y\left(x+1\right)-9\left(x+1\right)\)
=\(\left(x+1\right)\left(y-9\right)\)