Phân tích đa thức sau thành nhân tử:
x8+3x4+4
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a) \(=6x^2y^2\left(6xy-7\right)\)
b) \(=3xy\left(x^3y+5x-6\right)\)
c) \(=\left(ax+ab\right)-\left(bx+x^2\right)=a\left(b+x\right)-x\left(b+x\right)=\left(a-x\right)\left(b+x\right)\)
d) \(=3\left(2x-1\right)-\left(2x-1\right)^2=\left(2x-1\right)\left(3-2x+1\right)=\left(2x-1\right)\left(4-2x\right)=2\left(2x-1\right)\left(2-x\right)\)
\(a,=6x^2y^2\left(6xy-7\right)\\ b,=3xy\left(x^3y+5x-6\right)\\ c,=x\left(a-x\right)-b\left(a-x\right)=\left(x-b\right)\left(a-x\right)\\ d,=3\left(2x-1\right)-\left(2x-1\right)^2=\left(2x-1\right)\left(3-2x+1\right)=2\left(2-x\right)\left(2x-1\right)\)
Lời giải:
a.
$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$
$=(x^4+1-x^2)[(x^2+1)^2-x^2]$
$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$
b.
$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$
$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$
c.
$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$
$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$
$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$
c.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
a) \(2x^2+5x+2\)
\(=2x^2+4x+x+2\)
\(=2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(2x+1\right)\)
b) \(4x^2-4x-9y^2+12y-3\)
\(=\left(4x^2-4x+1\right)-\left(9y^2-12y+4\right)\)
\(=\left(2x-1\right)^2-\left(3y-2\right)^2\)
\(=\left(2x-1+3y-2\right)\left(2x-1-3y+2\right)\)
\(=\left(2x+3y-3\right)\left(2x-3y+1\right)\)
c) \(x^4-2x^3-4x^2+4x-3\)
\(=x^4+x^3-x^2+x-3x^2-3x+3x-3\)
\(=\left(x^4+x^3-x^2+x\right)-\left(3x^2+3x-3x+3\right)\)
\(=x\left(x^3+x^2-x+1\right)-3\left(x^3+x^2-x+1\right)\)
\(=\left(x^3+x^2-x+1\right)\left(x-3\right)\)
d) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(a,=y\left(y-2\right)\\ b,=3x\left(x^2-2x+1\right)=3x\left(x-1\right)^2\\ c,=\left(y-1\right)\left(27x^2+9x^3\right)=9x^2\left(x+3\right)\left(y-1\right)\\ d,=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\\ e,=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\\ f,=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\\ g,=\left(2-x\right)\left(x+1\right)\\ h,=\left(x-1\right)\left(3x-6\right)=3\left(x-1\right)\left(x-2\right)\)
a: =y(y-2)
b: \(=3x^2\left(x^2-2x+1\right)=3x^2\left(x-1\right)^2\)
d: \(=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
x4 + 4
= (x2)2 + 22
= x4 + 2.x2.2 + 4 – 4x2
(Thêm bớt 2.x2.2 để có HĐT (1))
= (x2 + 2)2 – (2x)2
(Xuất hiện HĐT (3))
= (x2 + 2 – 2x)(x2 + 2 + 2x)
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
\(=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
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: \(5x^2-10xy+5y^2-20\)
\(=5\left(x^2-2xy+y^2-4\right)\)
\(=5\left(x-y-2\right)\left(x-y+2\right)\)
x8+3x4+4
= x8+4x4+4-x4
= (x4+2)2-(x2)2
= (x4+2-x2).(x4+2+x2)
\(x^8+3x^4+4=(x^8+3x^4)+4=x^4(x^2+3)+4=(x^2+3)(x^4+4)\)