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19 tháng 10 2020

\(a,6x^3-9x^2=3x^2\left(2x-3\right)\)

\(b,4x^2y-8xy^2+10x^2y^2=2xy\left(2x-4y+5xy\right)\)

\(c,20x^2y-12x^3=4x^2\left(5y-3x\right)\)

\(d,4xy^2+8xyz=4xy\left(y+2z\right)\)

19 tháng 10 2020

\(6x^3-9x^2=3x^2\times\left(2x-3\right)\)

10 tháng 7 2019

a) \(4x^2-81=\left(2x\right)^2-9^2=\left(2x-9\right)\left(2x+9\right)\)

b) \(3\left(x-y\right)+5x\left(y-x\right)=3\left(x-y\right)-5x\left(x-y\right)=\left(x-y\right)\left(3-5x\right)\)

c) \(x\left(x+y\right)+6x+6y=x\left(x+y\right)+6\left(x+y\right)=\left(x+y\right)\left(x+6\right)\)

d) \(20x-5y=5\left(4x-y\right)\)

e) \(=2xy\left(5xy-4y+2y\right)\)

g) \(4xy+8xyz=4xy\left(1+2z\right)\)

31 tháng 7 2016

a, 4x3 -12x2 + 9x

=x(4x2 -12x + 9)

=x((2x)2 - 2.3.2x + 32)

=x(2x - 3)2

b,ab + c2 -ac - bc

=(ab - ac) + (c2 - bc)

=a(b - c) + c(c - b)

=a(b - c) - c(b - c)

=(a - c)(b - c)

c,4x2 - y2 + 1 - 4x
=((2x)2 - 2.2x + 1) - y2

=(2x - 1)2 - y2

=(2x - y -1)(2x + y - 1) 

d,6x- 7x - 20

= -(-6x2 + 7x + 20)

= -(-6x2 + 11x +10 + 10 - 4x)

= -((3x + 2)(-2x + 5) + 10 - 4x)

= -(3x + 2)(-2x + 5) -10 + 4x

= -(3x + 2)(-2x + 5) - 2(-2x + 5) 

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

\(a)\)

\(4x^2-y^2+2x+y\)

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

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

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

\(b)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x-9\right)\)

\(=\left(x-3\right)\left(x^2+5-9\right)\)

\(c)\)

\(12x^3+4x^2-27x-9\)

\(=\left(12x^3+4x^2\right)-\left(27x+9\right)\)

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

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

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

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

\(d)\)

\(16x^2+4x-y^2+y^2\)

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

\(4x\left(4x+1\right)\)

25 tháng 7 2017

Bài 1 : 

a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)

b)  \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)

c)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)

d)   \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)

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

BÀi 2 : 

a)   \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)

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

b)   \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)

\(=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)\)

c)  \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)

\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)

\(=\left(b+c-a\right)\left(d-c^2\right)\)

BÀi 3 : 

a)  \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)

b)  \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)

c)   \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)

d)   \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\)  \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)

Bài làm

a) 2x2y - 4xy2 + 6xy

= 2xy( x - 2y + 3 )

b) 4x3y2 - 8x2y+ 2x4y

= 2x2y( 2xy - 4y2 + x2 )

c) 9x2y- 3x4y- 6x3y+ 18y4 

= 3y2( 3x2y - x4 - 2x3 + 6y2 )

d) 7x2y- 21xy2z + 7xyz - 14xy

= 7xy( xy - 3yz + z - 2 )

# Học tốt #

30 tháng 10 2019

Ta có:

a) 6x2y - 3y2 - 2x2 + y = (6x2y - 2x2) - (3y2 - y) = 2x2(3y - 1) - y(3y - 1) = (2x2 - y)(3y - 1)

b)  2x2 + x - 4xy - 2y + 2x + 1 = (x2 + x) - (4xy + 2y) + (x2 + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)2

 = (x + x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1 - 2y)

c) 16x2y - 4xy2 - 4x3 + x2y = 4xy(4x - y) - x2(4x - y) = (4xy - x2)(4x - y)

d) 4x2 - 20x + 25 - 36y2 = (2x  - 5)2 - (6y)2 = (2x - 5 - 6y)(2x  - 5 + 6y)

e) x2 - 4y2 + 6x - 4y + 8 = (x2 + 6x + 9) - (4y2 + 4y + 1) = (x + 3)2 - (2y + 1)2 = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)

30 tháng 10 2019

g) Ta có : x10 + x5 + 1

= (x10 - x) + (x5 - x2) + (x2 + x + 1)

= x(x9 - 1) + x2(x3 - 1) + (x2 + x + 1)

= x(x3 - 1)(x6 + x3 + 1) + x2(x3 - 1) + (x2 + x + 1)

= (x7 + x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)(x2 + x + 1) + (x2 + x + 1)

= (x2 + x + 1)(x8 - x7 + x 5 - x4 + x2 - x + x4 + x3 + x2 + 1)

= (x2 + x + 1)(x8 - x7 + x5 + x3 - x + 1)

h) TT trên (dài dòng ktl)

28 tháng 12 2016

a) \(2x^2\left\{x^2+5x+6\right\}\)=\(2x^4+10x^3+12x^2\)

b) \(15x^2y^4:10x^2y\)=\(\frac{3}{2}y^3\)

c) \(\left\{16x^3y^2+20x^2y^3-8xy\right\}:4xy\)=\(4x^2y+5xy^2-2\)