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17 tháng 7 2019

(x^3-3x-3)(x^6+3x^4-6x^3+9x^2-9x+9)

2 tháng 3 2018

\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right).\)

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

vậy................

2 tháng 3 2018

\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right)\)

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

Vậy ........

Ta có: \(4x^2+12x+9\)

\(=4x^2+6x+6x+9\)

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

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

AH
Akai Haruma
Giáo viên
12 tháng 7 2021

Lời giải:
$x^4y^4-z^4=(x^2y^2)^2-(z^2)^2=(x^2y^2-z^2)(x^2y^2+z^2)$

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

$(x+y+z)^2-4z^2=(x+y+z)^2-(2z)^2=(x+y+z-2z)(x+y+z+2z)$
$=(x+y-z)(x+y+3z)$

$\frac{-1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2=\frac{-4x^2+12xy-9y^2}{36}$

$=-\frac{4x^2-12xy+9y^2}{36}=-\frac{(2x-3y)^2}{36}=-\left(\frac{2x-3y}{6}\right)^2$

12 tháng 7 2021

Câu trả lời của cô quá đúng luôn đấy

5 tháng 6 2016

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

5 tháng 6 2016

x2 + x - 6 

= x2 - 2x + 3x - 6

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

= ( x - 2 ) ( x + 3 )

31 tháng 7 2016

=m^3-3m^2-3m^2+9n+2m-6

=m^2(m-3)-3m(m3)+2(m-3)

=(m-3)(m^2-3m+2)=(m-3)(m^2-m-2m+2)

=(m-3)[m(m-1)-2(m-1)]

=(m-3)(m-2)(m-1)

\(m^3-6m^2+11m-6\)

\(m^3-6m^2+11m-6\)

\(=\left(m-1\right)\left(m-3\right)\left(m-2\right)\)

19) Ta có: \(-x^2-4x-4\)

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

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

20) Ta có: \(-4x^2-12x-9\)

\(=-\left(4x^2+12x+9\right)\)

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

21) Ta có: \(-4x^2-4x-1\)

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

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

22) Ta có: \(-x^2+6x-9\)

\(=-\left(x^2-6x+9\right)\)

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

23) Ta có: \(-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

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

24) Ta có: \(-x^2+8x-16\)

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

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

25) Ta có: \(-4x^2+12x-9\)

\(=-\left(4x^2-12x+9\right)\)

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

26) Ta có: \(a^2-a+b-b^2\)

\(=\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-1\right)\)

13) Ta có: \(y^2-2xy+2x-y\)

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

\(=\left(y-2x\right)\left(y-1\right)\)

14) Ta có: \(x-2xy+4y-2\)

\(=x\left(1-2y\right)-2\left(1-2y\right)\)

\(=\left(1-2y\right)\left(x-2\right)\)

15) Ta có: \(x^2-2xy+x-2y\)

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

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

16) Ta có: \(xy-z-y+xz\)

\(=x\left(y+z\right)-\left(y+z\right)\)

\(=\left(y+z\right)\left(x-1\right)\)

17) Ta có: \(2xy+3z-6y-xz\)

\(=\left(2xy-xz\right)+\left(3z-6y\right)\)

\(=x\left(2y-z\right)-3\left(2y-z\right)\)

\(=\left(2y-z\right)\left(x-3\right)\)

18) Ta có: \(2xy-2z+4y-xz\)

\(=\left(2xy+4y\right)+\left(xz+2z\right)\)

\(=2y\left(x+2\right)+z\left(x+2\right)\)

\(=\left(x+2\right)\left(2y+z\right)\)

26) Ta có: \(x^4-20x^2+64\)

\(=x^4-16x^2-4x^2+64\)

\(=x^2\left(x^2-16\right)-4\left(x^2-16\right)\)

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

27) Ta có: \(4x^3+6x^2+3x+1\)

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

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

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

28) Ta có: \(x^3-6x^2+12x-9\)

\(=x^3-3x^2-3x^2+9x+3x-9\)

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

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

29: Ta có: \(x^4+x^2+1\)

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

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

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

9 tháng 8 2021

26) Ta có: x4−20x2+64x4−20x2+64

=x4−16x2−4x2+64=x4−16x2−4x2+64

=x2(x2−16)−4(x2−16)=x2(x2−16)−4(x2−16)

=(x−4)(x+4)(x−2)(x+2)=(x−4)(x+4)(x−2)(x+2)

27) Ta có: 4x3+6x2+3x+14x3+6x2+3x+1

=4x3+4x2+2x2+2x+x+1=4x3+4x2+2x2+2x+x+1

=4x2(x+1)+2x(x+1)+(x+1)=4x2(x+1)+2x(x+1)+(x+1)

=(x+1)(4x2+2x+1)=(x+1)(4x2+2x+1)

28) Ta có: x3−6x2+12x−9x3−6x2+12x−9

=x3−3x2−3x2+9x+3x−9=x3−3x2−3x2+9x+3x−9

=x2⋅(x−3)−3x(x−3)+3(x−3)=x2⋅(x−3)−3x(x−3)+3(x−3)

=(x−3)(x2−3x+3)=(x−3)(x2−3x+3)

29: Ta có: x4+x2+1x4+x2+1

=x4+2x2+1−x2=x4+2x2+1−x2

=(x2+1)2−x2=(x2+1)2−x2

=(x2−x+1)(x2+x+1)

5 tháng 7 2017

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

c) \(36-12x+x^2=x^2-12x+36=x^2-6x-6x+36\)

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

\(x^4-y^4\)

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

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

\(4x^2+12x+9\)

\(=\left(2x\right)^2+2.2x.3+9\)

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

\(36-12x+x^2\)

\(=6^2-2.6.x+x^2\)

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