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25 tháng 2 2021

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

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

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

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

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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

 

15 tháng 12 2021

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)

6 tháng 10 2021

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

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

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

22 tháng 7 2023

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

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

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)

\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)

\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)

=1/5-1=-4/5

\(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)

d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)

\(=20x^3-30x^2+15x+4\)

\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)

8 tháng 8 2017

a) \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

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

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

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

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

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

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

\(=\dfrac{x}{x+y}\)

d) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\)

\(=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a-b+c\right)\left(a+b+c\right)}\)

\(=\dfrac{a+b-c}{a-b+c}\)

e) \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

\(=\dfrac{2x^2-x-15}{3x^2-10x+3}\)

\(=\dfrac{\left(x-3\right)\left(2x+5\right)}{\left(x-3\right)\left(3x-1\right)}\)

\(=\dfrac{2x+5}{3x-1}\)

8 tháng 8 2017

You're welcome :)) :)) :)) :)) :)) :)) :))

28 tháng 6 2017

Rút gọn phân thức

6 tháng 11 2018

a) \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)

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

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

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

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

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

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

Mình không chắc là đúng hoàn toàn nha!

6 tháng 11 2018

b) \(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)

= \(\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)

= \(\dfrac{x^3-6x^2y}{x-6y}\)

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

= \(x^2\)