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3 tháng 7 2017

a, \(\left(x^2-y^2\right)-\left(5x+5y\right)\)

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

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

b, \(5x^3-5x^2y-10x^2+10xy\)

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

\(=\left(5x-10x\right)\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)

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

f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)

\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)

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

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

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

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

e, \(3x^2-6xy+3y^2-12z^2\)

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

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

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

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

h, \(3x^2-3y^2-12x+12y=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+3y-12\right)\)

\(=3\left(x-y\right)\left(x+y-4\right)\)

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

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

3 tháng 7 2017

a. \(\left(x^2-y^2\right)-5\left(x+y\right)\)

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

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

b. \(5x^3-5x^2y-10x^2+10xy\)

\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)

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

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

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

d. \(x^3-3x^2+1-3x\)

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

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

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

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

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

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

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

e. \(3x^2-6xy+3y^2-12z^2\)

\(=3\left[x^2-2xy+y^2-4z^2\right]\)

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

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

f. \(3x^2-7x-10\)

\(=3x^2-7x-7-3\)

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

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

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

\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)

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

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

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

h. \(3x^2-3y^2-12x+12y\)

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

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)

\(=\left(x-y\right).3.\left(x+y-4\right)\)

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

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

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

P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )

16 tháng 10 2014

a) =(x-y)*(x+y)-(5*(x+y))

=(x+y)*(x-y-5)

Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung 

bai nao khong hieu thi pan nhan tin vào nick minh minh se giai đùm ban

17 tháng 10 2014

a) (x2 - y2) - 5(x + y)

= (x - y)(x + y) - 5 (x + y)

= (x + y) (x - y -5)

b) 5x3 - 5x2y - 10x2 + 10 xy

= 5[(x3 - x2y) - (2x2 - 2 xy)]

=5[x2(x - y) - 2x(x - y)]

=5x(x-y)(x - 2)

c) 2x2 - 5x = x(2x - 5)

d) x3 - 3x2 +1 - 3x 

= (x3 + 1) - (3x2 + 3x)

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

= (x + 1) [x2 - x + 1 - 3x]

= (x + 1)[x2 - 4x + 1]

= (x + 1)[x2 - 2.x.2 + 22 - 22 + 1]

= (x + 1)[(x - 2)2 - 3]

\(\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)

e) 3x2 - 6xy + 3y2 - 12z2

= 3[ x2 - 2xy + y2 - 4z2]

= 3[ (x - y)2 - (2z)2]

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

f) 3x2 - 7x - 10

= 3x2 - 7x - 7 - 3

= (3x2 -3) - (7x + 7)

= 3(x- 1) - 7(x + 1)

= 3 (x + 1)(x - 1) - 7(x + 1)

= (x + 1)[3(x - 1) - 7]

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

g) x4 + 1 - 2x2 = (x2)2 - 2.x2 + 1 = (x- 1)2

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

h) 3x2 - 3y2 - 12x + 12y

= 3(x- y2) - 12(x - y)

= 3(x - y)(x + y) - 12(x -y)

= (x - y) [3(x + y) - 12]

= (x - y). 3. (x+y - 4)

j) x2 - 3x + 2 = x2 - x - 2x +2

= x(x - 1) - 2(x -1)

=(x - 1)(x - 2)

15 tháng 10 2017

a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)

b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y

c,đề bài thiếu  ẩn ở hạng tử thứ nhất ạ !

15 tháng 10 2017

b mình viết đúng rồi mà, c hạng tử 1 là x^3

8 tháng 9 2021

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

17 tháng 5 2019

A= 3xy-11x2-5y.8xy-5+6

=(3-11-5.8-5+6).(x2.x2.x).(y.y.y)

=-47x5y3

24 tháng 7 2018

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

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

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

f) \(x^2+2xy+y^2-2x-2y\)

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

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

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

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

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

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

24 tháng 7 2018

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

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

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

f) \(x^2+2xy+y^2-2x-2y\)

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

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

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

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

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

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

6 tháng 4 2020

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)

6 tháng 4 2020

Thank you

22 tháng 12 2023

Bài 2:

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

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

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

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

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

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

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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