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1 tháng 11 2018

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

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

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

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

1 tháng 11 2018

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

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

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

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

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

hok tốt ...

3 tháng 3 2020

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

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

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

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

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

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

HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023

\(\frac{{9{x^2} + 5x + x}}{{3x}} = \frac{{9{x^2} + 6x}}{{3x}} = \frac{{9{x^2}}}{{3x}} + \frac{{6x}}{{3x}} = 3x + 2\)

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

1 tháng 12 2018

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

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

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

9 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}=\frac{1}{x-3}\)

14 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x-14}=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}\)

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