K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 11 2017

cho mik sửa lại câu

b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

29 tháng 11 2017

b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)}{3x+2y}-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)-\left(6xy+2y\right)+\left(2y-9x^2\right)}{3x+2y}\)

\(=\dfrac{6xy+4y^2-6xy-2y+2y-9x^2}{3x+2y}\)

\(=\dfrac{4y^2-9x^2}{3x+2y}\)

\(=\dfrac{-\left(9x^2-4y^2\right)}{3x+2y}\)

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

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

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

\(=-3x+2y\)

7 tháng 8 2021

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

1 tháng 12 2021

\(a,\left(12x^2y^2-6xy^2\right):3xy+2y=6xy^2\left(2x-1\right):3xy+2y=2y\left(2x-1\right)+2y=4xy-2y+2y=4xy\)

\(b,\dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{x-1}\)

\(c,\dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} \)

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

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

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

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

\(=\dfrac{2}{x+1}\)

 

1 tháng 12 2021

\(a,=4xy-2y+2y=4xy\\ b,\dfrac{4}{x+1}+\dfrac{8}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ c,\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}=\dfrac{x-1-x-1+2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+1}\)

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

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

16 tháng 12 2022

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

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

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

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

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

24 tháng 6 2017

Phân thức đại số

Phân thức đại số

NV
30 tháng 7 2021

a.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)

\(\Rightarrow3x+2=2x\left(x+y\right)+y\)

\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)

\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)

Thế vào pt đầu ...

Câu b chắc chắn đề sai

19 tháng 12 2021

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)