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a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
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.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
\(a,=5x\left(4x-1\right)\\ b,=y^2-\left(x-1\right)^2=\left(y-x+1\right)\left(y+x-1\right)\\ c,=6x^2+3x-4x-2=3x\left(x+2\right)-2\left(x+2\right)=\left(3x-2\right)\left(x+2\right)\)
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
Bài 1:
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)
1, =3x (2x -3y)
c, = 3x(x-y) -2(x-y)
= (3x-2)(x-y)
2, Ta có: x2 -6x+10= (x-3)2 +11
Nhận xét: (x-3)2 >= 0 với mọi số thực x
=> (x-3)2 +1 >= 1 >0 (đpcm)
`a)x^3-8x^2+16x`
`=x(x^2-8x+16)`
`=x(x-4)^2`
`b)x^2+4y^2+2x-4y-4xy-24`
`=(x-2y)^2+2(x-2y)-24`
`=(x-2y)^2-4(x-2y)+6(x-2y)-24`
`=(x-2y-4)(x-2y+6)`
`c)x^4+x^3-x^2-2x-2`
`=x^4-2x^2+x^3-2x+x^2-2`
`=x^2(x^2-2)+x(x^2-2)+x^2-2`
`=(x^2-2)(x^2+x+1)`
c: \(x^2-10x+21=\left(x-3\right)\left(x-7\right)\)
a: \(x^2y+xy^3-xy-y^3\)
\(=xy\left(x-1\right)+y^3\left(x-1\right)\)
\(=y\left(x-1\right)\left(x+y^2\right)\)
\(a) x^2y+xy^3-xy-y^3\\=(x^2y+xy^3)-(xy+y^3)\\=xy(x+y^2)-y(x+y^2)\\=(x+y^2)(xy-y)\\=y(x+y^2)(x-1)\\b)2x^2+5x+8(xem lại đề)\\c)x^2-10x+21\\=x^2-3x-7x+21\\=x(x-3)-7(x-3)\\=(x-3)(x-7)\)
\(a,=xy\left(x+y^2\right)-y\left(x+y^2\right)=y\left(x+y^2\right)\left(x-1\right)\\ c,=x^2-7x-3x+21=\left(x-7\right)\left(x-3\right)\)
A = x8 + 2x5 - 2x4 + x2 - 2x - 100 + 10x.(x4 + x) + (5x - 1)2
A = (x8 + 2x5 + x2) - (2x4 + 2x) + 10x.(x4 + x) + (5x - 1)2 - 100
A = (x4 + x)2 - 2(x4 + x) + 10x. (x4 + x) + (5x -1)2 - 100
A = (x4 + x)2 + (x4 + x).(10x - 2) + (5x - 1)2 - 100
A = [(x4 + x)2 + 2.(x4 + x).(5x - 1) + (5x - 1)2 ] - 100
A = [x4 + x + 5x - 1]2 - 102
A = (x4 + 6x - 11).(x4 + 6x + 9)
Hok tốt ^_^