Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)
\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)
\(=\frac{3x-2}{9x^2-4}\)
\(=\frac{1}{3x+2}\)
\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)
\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)
\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)
\(=\frac{\left(-x^2-3\right)\left(x-3\right)}{\left(x-3^2\right)\left(x+3\right)}\)
\(=\frac{-x^2-3}{\left(x-3\right)\left(x+3\right)}\)
học tốt
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
\(a,\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3\left(x-2\right)}{\left(3x+2\right)\left(3x-2\right)}\)
\(=\frac{3x+2-\left(3x-2\right)+3\left(x-2\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{3x+2}\)
\(b,\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)
\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}-\frac{3}{\left(x-3\right)\left(x-3\right)}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{18-3x-9-x^2+3x}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-x^2+9}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}=-\frac{1}{x-3}\)