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16 tháng 12 2018

Câu 1:

a/ (-5x3)(2x2+3x-5)

=-10x5-15x4+25x3

b/(2x-1)x

=2x2-x

c/(x-y)(3x2+4xy)

=3x3+4x2y-3x2y-4xy2

=3x3 +x2y-4xy2

Câu 2:

a/ x3-2x2+x

=x(x2-2x+1)

=x(x-1)2

b/x2-x-12

=x2 +3x-4x-12

=(x2 +3x)+(-4x-12)

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

=(x+3)(x-4)

c/ 2x-6

=2(x-3)

e/ x2+4x+4-y2

=(x2+4x+4)-y2

=(x+2)2-y2

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

d/ x2-2xy+y2-16

=(x2-2xy+y2)-16

=(x-y)2-16

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

Câu 3:

a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)

d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)

e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)

3 tháng 5 2020
https://i.imgur.com/88Zm20M.jpg
3 tháng 5 2020
https://i.imgur.com/zsGzAKT.jpg
27 tháng 12 2017

Mấy thánh đăng sớm z

27 tháng 12 2017

a.

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

b.https://hoc24.vn/hoi-dap/question/531769.html

Mình làm bài này r ; mà nó dài nên k mún làm lại

c.

\(5x^2-10xy+5y^2-20z^2\\ =5\left(x^2-2xy+y^2-4z^2\right)\\ =5\left(\left(x-y\right)^2-4z^2\right)\\ =5\left(x-y-2z\right)\left(x-y+2z\right)\)

m)

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

3 tháng 9 2018

pạn ơi pạn đã lm đk chưa? nếu lm đk oy cho mk xem cách lm bài 2 nhé. cảm ơn pạn nhìu lắm

16 tháng 8 2018

a) Sửa đề

\(x^4+2x^3+x^2\)

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

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

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

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

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

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

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

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

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

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

c) \(5x^2-10xy+5y^2-20z^2\)

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

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

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

3 tháng 10 2017

đề bài đâu

ucche

3 tháng 10 2017

cô hk ghi nha bn

sorry nha

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 ^^