K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 9 2019

\(a.a\left(m-n\right)+m-n\)

\(=a\left(m-n\right)+\left(m-n\right)\)

\(=\left(a+1\right)\left(m-n\right)\)

\(b.ma+mb-a-b\)

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

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

\(c.4x+by+4y+bx\)

\(=\left(4x+4y\right)+\left(bx+by\right)\)

\(=4\left(x+y\right)+b\left(x+y\right)\)

\(=\left(b+4\right)\left(x+y\right)\)

\(d.1-ax-x+a\)

\(=\left(a-ax\right)+\left(1-x\right)\)

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

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

4 tháng 9 2019

1.a(m-n)+m-n=am-an+m-n=(am+m)-(an+n)=m(a+1)-n(a+1)=(a+1)(m-n)

2.ma+mb-a-b=(ma-a)+(mb-b)=a(m-1)+b(m-1)=(m-1)(a+b)

3.4x+by+4y+bx=(4x+bx)+(4y+by)=x(4+b)+y(4+b)=(4+b)(x+y)

4.1-ax-x+a=(1+a)-(ax+x)=(1+a)-x(a+1)=(1+a)(1-x)

16 tháng 7 2017

1, \(a.\left(b+c\right)+3b+3c=a.\left(b+c\right)+3.\left(b+c\right)\)\(\left(b+c\right).\left(3+a\right)\)

2, \(a.\left(m-n\right)+\left(m-n\right)=\left(m-n\right).\left(a+1\right)\)

3, \(7a^2-7ax-9a+9x=7a.\left(a-x\right)-9.\left(a-x\right)\)\(\left(a-x\right).\left(7a-9\right)\)

4, \(4x+by+4y+bx=4.\left(x+y\right)+b.\left(x+y\right)\)\(\left(x+y\right).\left(4+b\right)\)

5, \(ay-ax-2x+2y=a.\left(y-x\right)+2.\left(y-x\right)\)\(\left(y-x\right).\left(a+2\right)\)

    Chúc bạn học tốt. Có gì không hiểu thì chat hỏi mik nhé. ^^

16 tháng 7 2017

thank bn nhiều nha

18 tháng 7 2017

a, (4x+4y)+(by+bx)= 4(x+y)+b(x+y)=(x+y)(4+b)

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

c, (3ax-2bx)-(6ay-4by)= x(3a-2b)-2y(3a-2b)=(3a-2b)(x-2y)

d, (ma+na-pa)-(mb+nb-pb)= a(m+n+p)-b(m+n-p)=(m+n+p)(a-b)

18 tháng 7 2017

a) 4x+bx+by+4y                b)2x2+xy-2x-y                c)3ax-2bx-6ay+4by                          d)ma-mb+na-nb-pa+pb

=x(4+b)+y(b+4)                 =2x(x-1)+y(x-1)              =3ax-6ay-2bx+4by                            =m(a-b)+n(a-b)-p(a-b)

=(x+y)(b+4)                      =(x-1)(2x+1)                   =3a(x-2y)-2b(x-2y)=(3a-2b)(x-2y)        =(a-b)(m+n-p)

6 tháng 8 2019

\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)

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

\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)

\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)

\(=\left(4x-y\right)\left(a+b-c-1\right)\)

\(c,x^k+1-x^k-1\)

\(=0?!?!\)

\(d,x^m+3-x^m+1\)

\(=4\)

\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)

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

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

6 tháng 8 2019

\(f,81a^2+18a+1\)

\(=\left(9a\right)^2+2.9a+1\)

\(=\left(9a+1\right)^2\)

\(g,25a^2.b^2-16c^2\)

\(=\left(5ab\right)^2-\left(4c\right)^2\)

\(=\left(5ab+4c\right)\left(5ab-4c\right)\)

\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)

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

\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)

\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)

\(=2by.2ax\)

\(=4axby\)

24 tháng 10 2021

\(a,=x^2-mx-nx+mn=x\left(x-m\right)-n\left(x-m\right)=\left(x-n\right)\left(x-m\right)\\ b,=a\left(x-y\right)-b\left(x-y\right)+\left(a-b\right)\\ =\left(x-y\right)\left(a-b\right)+\left(a-b\right)=\left(a-b\right)\left(x-y+1\right)\)

24 tháng 10 2021

b: \(=a\left(x-y\right)-b\left(x-y\right)+a-b\)

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

12 tháng 9 2015

a)x^2+2x-4y^2-4y

=(x2-4y2)+(2x-4y)

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

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

b)x^4-6x^3+54x-81

=(x4-81)+(-6x3+54x)

=(x2-9)(x2+9)-6x.(x2-9)

=(x2-9)(x2+9-6x)

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

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

c)ax^2+ax-bx^2-bx-a+b

=(ax2-bx2)+(ax-bx)+(-a+b)

=x2.(a-b)+x.(a-b)-(a-b)

=(a-b)(x2+x+1)

 

 

25 tháng 7 2018

1. ac - ad + (c - d) = a(c - d) + (c - d) = (a + 1)(c - d)

2. ax + ay - x - y = a(x + y) - (x + y) = (a - 1)(x + y)

3. 4x + by + 4y + bx = 4(x + y) + b(x + y) = (b + 4)(x + y)

4. 1 - ax - x + a = a(1 - x) + (1 - x) = (a + 1)(1 - x)

25 tháng 7 2018

1. ac-ad +(c-d)

= a(c-d) +(c-d)

=(c-d) (a+1)

2. ax+ay-x-y

=(ax+ay)-(x+y)

= a(x+y)-(x+y)

=(x+y) (a-1)

3. 4x +by +4y+bx

=(4x+4y) +(bx+by)

=4(x+y) +b(x+y)

=(4+b) (x+y)

4.=(1+a)-(x+ax)

=(1+a)-x(1+a)

=(1+a)(1-x)

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

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

30 tháng 7 2021

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

n, \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-\left(2z\right)^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)

23 tháng 10 2021

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)