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29 tháng 7 2021

Ta có:(x2-y2)\(.\dfrac{x^2+y^2}{y^4-x^2y^2}\)\(=\left(x^2-y^2\right).\dfrac{x^2+y^2}{y^2\left(y^2-x^2\right)}=-\dfrac{x^2+y^2}{y^2}\)

Ta có:\(\dfrac{4x^2-9y^2}{xy}:\left(2x-3y\right)=\dfrac{\left(2x-3y\right)\left(2x+3y\right)}{xy}.\dfrac{1}{\left(2x-3y\right)}=\dfrac{2x+3y}{xy}\)

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

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

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

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

2 tháng 7 2018

a/ \(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=-\dfrac{4x+2}{x\left(1-3x\right)}\cdot\dfrac{1-3x}{x^2+3x}=-\dfrac{4x^2+2}{x\left(x^2+3x\right)}\)

b/ \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}=-\dfrac{2\left(2x+3y\right)}{1-x}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(2x+3y\right)^2}=\dfrac{-2\left(x+1\right)}{2x+3y}=\dfrac{-2x-2}{2x+3y}\)

c/ \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}=\dfrac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}\cdot\dfrac{2x+y}{x\left(x^2+xy+y^2\right)}=\dfrac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y}\cdot\dfrac{1}{x\left(x^2+xy+y^2\right)}=\dfrac{x-y}{y}\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

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

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

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

Bài 4:

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a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y

=5x^3-7x^2y+2xy^2+5x-2y

b: =(x^2-1)(x+2)

=x^3+2x^2-x-2

c: =1/2x^2y^2(4x^2-y^2)

=2x^4y^2-1/2x^2y^4

d: =(x^2-1/4)(4x-1)

=4x^3-x^2-x+1/4

e: =x^2-2x-35+(2x+1)(x-3)

=x^2-2x-35+2x^2-6x+x-3

=3x^2-7x-38

26 tháng 12 2021

c: \(=x^2+6xy+9y^2\)

e: \(=x^4-4y^2\)

6 tháng 1 2021

B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)

b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)

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

B2:

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

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

Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

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

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

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)