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23 tháng 12 2020

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

b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

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

21 tháng 12 2020

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

\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{25-x^2}=\frac{4}{x-5}-\frac{1}{x+5}+\frac{x^2-13x}{x^2-25}\)

\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{4x+20-x+5+x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)

19 tháng 12 2018

Thực hiện phép tính

a,  6x3y5z : 3xy3z=2x2y2

b,  \(\frac{3x+6}{x+2}+\frac{2x+4}{x+2}\)

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

=3+2=5

15 tháng 12 2020

\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{4x-x^2}{4-x^2}\)

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

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

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

15 tháng 12 2020

\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{4x-x^2}{4-x^2}\)

\(=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2-4x}{x^2-4}\)

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

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

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

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