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8 tháng 10 2017

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8 tháng 10 2017

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

\(2,4ax^3-ax=ax\left(4x^2-1\right)=ax\left[\left(2x\right)^2-1^2\right]\) \(=ax\left(2x-1\right)\left(2x+1\right)\)

\(3,x^3-2x^2+x\)

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

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

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

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

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

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

\(=y\left(1^2-2.1.2x+\left(2x\right)^2\right)^{ }\)

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

22 tháng 7 2016

1not nhac/bai

1) = 3(x-y) +(x+y)(x-y) =(x-y)(x+y+3)

27 tháng 9 2020

a, x4 + 2x3 +x2 = x+x+x3 +x2  =(x4+x3 )+(x3 +x) =x3(x +1 ) + x(x+1 ) =(x+1)(x3+x2)

27 tháng 9 2020

a) x4 + 2x3 + x2

= x2(x2 + 2x + 1)

= x2(x + 1)2

= [x(x + 1)]2

= (x2 + x)2

b) 5x3 - 10xy + 5y2 - 20z2

= 5(x3 - 2xy + y2 - 4z2)

c) x2y - xy2 + x3 - y3

= xy(x - y) + (x - y)(x2 + xy + y2)

= (x - y)(x2 + 2xy + y2)

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

d) x2 - xy + 4x - 2y  + 4

= (x2 + 4x + 4) - (xy + 2y)

= (x + 2)2 - y(x + 2)

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

d) x2 - x - 6

= x2 - 3x + 2x - 6

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

= (x + 2)(x - 3)

f) 3x2 - 5x - 8

= 3x2 + 3x - 8x - 8

= 3x(x + 1) - 8(x + 1)

= (3x - 8)(x + 1)

g) x3 + 3x2 + 6x + 4

= (x3 + 3x2 + 3x + 1) + (3x + 3)

= (x + 1)3 + 3(x + 1)

= (x + 1)[(x + 1)2 + 3]

h) 3x3 - 5x2 - 6x + 8

= 3x3 - 3x2 - 2x2 - 6x + 8

= 3x3 - 3x2 - 2x2 + 2x - 8x + 8

= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)

= (3x2 - 2x - 8)(x - 1)

27 tháng 9 2020

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

b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

c) \(x^2y-xy^2+x^3-y^3\)

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

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

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

27 tháng 9 2020

d) \(x^2-xy+4x-2y+4\)

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

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

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

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

f) \(3x^2-5x-8\)

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

\(=3x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-8\right)\)

12 tháng 7 2019

a,\(xy+3x-7y-21\)

\(=x\left(y+3\right)-7\left(y+3\right)\)

\(=\left(y+3\right)\left(x-7\right)\)

12 tháng 7 2019

\(b,2xy-15-6x+5y\)

\(=\left(2xy-6x\right)+\left(-15+5y\right)\)

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

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

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

26 tháng 12 2018

1,4x2.(5x3+2x-1)

=4x2.5x3+4x2.2x-4x2.1

20x5+8x3-4x2

2,4x3y2:x2

=4xy2

3,(15x2y3-10x3y3+6xy):5xy

15x2y3:5xy-10x3y3:5xy+6xy:5xy

3xy2-2x2y2+\(\dfrac{6}{5}\)

26 tháng 12 2018

cảm ơn bạn nhé ^^

13 tháng 12 2022

1: \(=20x^5+8x^3-4x^2\)

2: \(=4xy^2\)

3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)

4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)

6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)

7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)

8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)

9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=4x^2-2x+\dfrac{1}{4}\)

10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)

\(=\dfrac{x^2-7}{2\left(x-1\right)}\)

12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

15:=x^3-y^3+2

13 tháng 12 2022

1: \(=20x^5+8x^3-4x^2\)

2: \(=4xy^2\)

3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)

4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)

6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)

7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)

8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)

9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=4x^2-2x+\dfrac{1}{4}\)

10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)

\(=\dfrac{x^2-7}{2\left(x-1\right)}\)

12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

15:=x^3-y^3+2

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

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