\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4y^2-9x^2}\)
K
Khách

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

1 tháng 3

Sửa đề:

\(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)

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

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

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

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

1 tháng 3

Olm chào em. Em cần làm gì với biểu thức này thfi ghi rõ yêu ra, em nhé. Như vậy em mới có thể nhận được sự trợ giúp tốt nhất từ cộng đồng.

11 tháng 12 2019

\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)

\(=\frac{-2}{x^2}\)

\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

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

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

\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

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

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

\(=\frac{x+3}{x+1}\)

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

20 tháng 12 2016

a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)

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

Chắc chắn đúng, mik nhaaaaaa

18 tháng 12 2017

\(\frac{3x^2+3x+3}{4x+4}\)\(\frac{9x^3-9}{2x^2-2}\)\(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\)\(\frac{9\left(x^3-1\right)}{2\left(x^2-1\right)}\)

\(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\)\(\frac{2\left(x-1\right)\left(x+1\right)}{9\left(x-1\right)\left(x^2+x+1\right)}\)\(\frac{1}{6}\)

18 tháng 12 2017
helpppp
3 tháng 12 2019

a) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

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

b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)

3 tháng 12 2019

c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)

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

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

e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

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

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

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

\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\) 

f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)

\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)

14 tháng 12 2018

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

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

14 tháng 12 2018

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

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

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)

3 tháng 3 2020

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

\(=\frac{-3}{x-3}\)

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)

\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)

\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)

chỗ cuối tớ sai 

\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)

đây nha , e xin lỗi