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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
MN
3 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\frac{\left(9+x^2-3x\right)\left(x+3\right)3x}{x\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)
\(=\frac{-3}{x-3}\)
PV
5
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 ...
KT
1
20 tháng 12 2016
a) \(\frac{3x}{2x+4}+\frac{x+3}{x^2-4}\)
\(=\frac{3x}{2\left(x+2\right)}+\frac{x+3}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)+2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{3x^2-6x+2x+6}{2\left(x^2-4\right)}\)
\(=\frac{3x^2-4x+6}{2\left(x^2-4\right)}\)
\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
\(=\frac{\left(x+3\right).\left(x-3\right)}{x.\left(x-3\right)}-\frac{x^2}{\left(x-3\right).x}+\frac{9}{x.\left(x-3\right)}=\frac{x^2-9-x^2+9}{x.\left(x-3\right)}=0\)