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10: \(x\left(x-y\right)+x^2-y^2\)

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

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

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

11: \(x^2-y^2+10x-10y\)

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

\(=\left(x-y\right)\left(x+y+10\right)\)

12: \(x^2-y^2+20x+20y\)

\(=\left(x^2-y^2\right)+\left(20x+20y\right)\)

\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+20\right)\)

13: \(4x^2-9y^2-4x-6y\)

\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)

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

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

14: \(x^3-y^3+7x^2-7y^2\)

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

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)

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

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

15: \(x^3+4x-\left(y^3+4y\right)\)

\(=x^3-y^3+4x-4y\)

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

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

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

16: \(x^3+y^3+2x+2y\)

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

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

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

17: \(x^3-y^3-2x^2y+2xy^2\)

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

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

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

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

18: \(x^3-4x^2+4x-xy^2\)

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

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

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

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

8 tháng 12 2023

Phân tích đa thức thành nhân tử nha

22 tháng 8 2018

\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)

                                                                  \(=2x^3+16x^2-5x\)

                                                                  \(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)

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

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

30 tháng 11 2017

\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)

\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)

\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)

\(=\left(2x-5\right).12\)

Những câu khác làm tương tự

26 tháng 12 2018

1) 4x\(^2\).(5x3+2x-1)

= 20x\(^5\)+8x\(^3\)-4x\(^2\).

2) 4x\(^3\): x2

= 4x

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

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

4) (5x3+14x2+12x+8 ): (x+2)

= 5x2+4x+4

5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)

=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)

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

7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)

= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).

8)\(\dfrac{1}{2}\)x2y2.(2x+y)(2x-y)

= \(\dfrac{1}{2}\)x2y2.(4x2-2xy+2xy-y2)

= \(\dfrac{1}{2}\)x2y2.(4x2-y2)

= 2x4y2-\(\dfrac{1}{2}\)x2y4

9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)

= x2.(4x-1)

= 4x3-x2

10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)

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

11) x2-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)

12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)

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

26 tháng 12 2018

cảm ơn bạn nhé ^^

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

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

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

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

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

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