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Bài 4:

a) Ta có: \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)

\(=\left(x^9-x^7\right)-\left(x^6-x^4\right)-\left(x^5-x^3\right)+\left(x^2-1\right)\)

\(=x^7\left(x^2-1\right)-x^4\left(x^2-1\right)-x^3\left(x^2-1\right)+\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^7-x^4-x^3+1\right)\)

\(=\left(x^2-1\right)\cdot\left[x^4\left(x^3-1\right)-\left(x^3-1\right)\right]\)

\(=\left(x^2-1\right)\cdot\left(x^3-1\right)\cdot\left(x^4-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\cdot\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)

\(=\left(x-1\right)^3\cdot\left(x+1\right)^2\cdot\left(x^2+1\right)\cdot\left(x^2+x+1\right)\)

5 tháng 8 2020

a, Ta có : \(x^5-x^4-x^3-x^2-x-2\)

\(=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)

\(=x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

\(=\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)\)

19 tháng 6 2016

a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1

=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1

=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)

=(x2+x+1)(x5-x4+x3-x+1)

b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1

=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)

=(2x2-6x+1)(2x2+6x+1)

c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)

d)3(x4+x2+1)-(x2+x+1)

=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2

=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)

=(x2+x+1)(3x2-3x+2)

e)bạn tự làm nhé

6 tháng 8 2020

a, ( x2 + x )2 - 14 ( x2 + x ) + 24

= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24

= (x2 + x).(x2 + x -2) - 12(x2 + x -2)

= (x2 + x -2).(x2 + x -12)

= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)

=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]

= (x+2).(x-1).(x+4).(x-3)

= x4 + 2x3 - 13x2 - 14x + 24

b, ( x2 + x )2 + 4x2 + 4x - 12

= x4 + 2x3 + x2 + 4x2 + 4x -12

= x4 + 2x3 + 5x2 + 4x -12

c, x4 + 2x3 + 5x2 + 4x - 12

= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12

= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)

= (x-1) . (x3 + 3x2 + 8x +12)

= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)

= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]

= (x-1).(x+2).(x2 + x+ 6)

Bài 6:

a) Ta có: \(x^2-4xy+4y^2-2x+4y-35\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)

\(=\left(x-2y\right)^2-7\cdot\left(x-2y\right)+5\left(x-2y\right)-35\)

\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)

\(=\left(x-2y-7\right)\left(x-2y+5\right)\)

b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x\right)^2+3\cdot\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\cdot\left(x^2+x\right)-10\)

\(=\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10\)

\(=\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)\)

\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)

\(=\left(x^2+x+5\right)\left(x-1\right)\left(x+2\right)\)

c) Ta có: \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384+16\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+400\)

\(=\left(x^2+10x+20\right)^2\)

d) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

e) Ta có: \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\)

\(=\left(x^2+10x\right)^2+16\left(x^2+10x\right)+8\left(x^2+10x\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+8\right)\)

\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)

AH
Akai Haruma
Giáo viên
5 tháng 8 2020

d)

$x^4+2x^3+2x^2+2x+1$

$=(x^4+2x^3+x^2)+(x^2+2x+1)$

$=(x^2+x)^2+(x+1)^2=x^2(x+1)^2+(x+1)^2$

$=(x+1)^2(x^2+1)$

e)

$x^2y+xy^2+x^2z+y^2z+2xyz$

$=xy(x+y)+z(x^2+y^2)+2xyz$

$=xy(x+y)+z(x^2+y^2+2xy)$

$=xy(x+y)+z(x+y)^2=(x+y)(xy+zx+zy)$

f)

$x^5+x^4+x^3+x^2+x+1$

$=(x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)$

$=(x+1)(x^4+x^2+1)$

$=(x+1)[(x^4+2x^2+1)-x^2]$

$=(x+1)[(x^2+1)^2-x^2]=(x+1)(x^2+1-x)(x^2+1+x)$

AH
Akai Haruma
Giáo viên
5 tháng 8 2020

a)

$x^4-2x^3+2x-1=(x^4-2x^3+x^2)-(x^2-2x+1)$

$=(x^2-x)^2-(x-1)^2$

$=x^2(x-1)^2-(x-1)^2=(x-1)^2(x^2-1)=(x-1)^2(x-1)(x+1)$

$=(x-1)^3(x+1)$

b)

$a^6-a^4+2a^3+2a^2$

$=a^4(a^2-1)+2a^2(a+1)$

$=a^4(a-1)(a+1)+2a^2(a+1)$

$=(a+1)[a^4(a-1)+2a^2]$

$=a^2(a+1)[a^2(a-1)+2]$

$=a^2(a+1)(a^3-a^2+2)=a^2(a+1)[a^2(a+1)-2(a^2-1)]$

$=a^2(a+1)[a^2(a+1)-2(a-1)(a+1)]$

$=a^2(a+1)(a+1)(a^2-2a+2)=a^2(a+1)^2(a^2-2a+2)$

c)

$x^4+x^3+2x^2+x+1$

$=(x^4+2x^2+1)+(x^3+x)$

$=(x^2+1)^2+x(x^2+1)=(x^2+1)(x^2+1+x)$

Bài 4:

a) Ta có: \(a^4+a^2+1\)

\(=a^4+2a^2+1-a^2\)

\(=\left(a^2+1\right)^2-a^2\)

\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)

b) Ta có: \(a^4+a^2-2\)

\(=a^4+2a^2-a^2-2\)

\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)

\(=\left(a^2+2\right)\left(a^2-1\right)\)

\(=\left(a^2+2\right)\left(a-1\right)\left(a+1\right)\)

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+5x^2-x^2-5\)

\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)

\(=\left(x^2+5\right)\left(x^2-1\right)\)

\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)

d) Ta có: \(x^3-19x-30\)

\(=x^3-25x+6x-30\)

\(=x\left(x^2-25\right)+6\left(x-5\right)\)

\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+5x+6\right)\)

\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)

\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

e) Ta có: \(x^3-7x-6\)

\(=x^3-4x-3x-6\)

\(=x\left(x^2-4\right)-3\left(x+2\right)\)

\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x-3\right)\)

\(=\left(x+2\right)\left(x^2-3x+x-3\right)\)

\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)

\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)

f) Ta có: \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-7x+2x-14\right)\)

\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)

\(=x\left(x-7\right)\left(x+2\right)\)