Phân tích đa thức thành nhân tử: x4 +5x3 + 10x - 4
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\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)
a: =x^4+2x^2+5x^3+10x-2x^2-4
=(x^2+2)(x^2+5x-2)
b; =(x+y)^3+z^3-3xy(x+y)-3xyz
=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1
=(x^2+x+1)(x^6-x^5+x^3-x^2+1)
\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)
\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
= x2(5x-7)+2(5x-7)
= (5x-7)(x2+2)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+3x+9+3x)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
=(5x3+10x)-(7x2+14)
=5x(x2+2)-7(x2+2)
=(x2+2)(5x-7)
\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)
\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
\(=5x\left(x^2-2xy+y^2\right)\)
\(=5x\left(x-y\right)^2\)
x4+4 = (x2)2+22 = x4 + 2.x2.2 + 4 – 4x2
= (x2 + 2)2 – (2x)2 = (x2-2x+2)(x2+2x+2)
Ta có: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
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)
x 4 - 5 x 2 + 4 = x 4 - 4 x 2 - x 2 + 4 = x 4 - 4 x 2 - x 2 - 4 = x 2 x 2 - 4 - x 2 - 4 = x 2 - 4 x 2 - 1 = x + 2 x - 2 x + 1 x - 1
\(a,=\left(x-2\right)\left(9x^2y^2-6x^3y^2\right)=3x^2y^2\left(3-2x\right)\left(x-2\right)\\ b,=5x\left(x^2-y^2\right)+20x\left(x+y\right)=5x\left(x-y\right)\left(x+y\right)+20x\left(x+y\right)\\ =5\left(x+y\right)\left(x^2-xy+4x\right)\\ c,=8x^2+2x-12x-3=2x\left(4x+1\right)-3\left(4x+1\right)=\left(2x-3\right)\left(4x+1\right)\)
d) x4 + 2x3 - 4x – 4 = (x4 – 4) + (2x3 – 4x) = (x2 – 2)(x2 + 2) + 2x(x2 – 2)
= (x2 – 2)(x2 + 2 + 2x) = (x - √2)( x + √2)( x2 + 2 + 2x)
x^4+ 2x^2 + 5x^3 + 10x- 2x^2 - 4
= x^2. [x^2+ 2 ] +5.x[x^2+2] - 2.[x^2+2]
= [x^2+5x-2] . [x^2+2]