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24 tháng 12 2020

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

\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)

b, Sua de :  \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

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

14 tháng 12 2020

\(\frac{x+2}{x}+\frac{2x-1}{2-x}-\frac{x-8}{x^2-2x}\)

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

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

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

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

21 tháng 12 2020

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

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

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

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

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

24 tháng 12 2020

a, \(\frac{x+1}{2x+6}=\frac{x+1}{2\left(x+3\right)}\)

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

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

c, \(\frac{x-x-2xy+x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x-2xy}{x+2y}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}\)

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

25 tháng 8 2021

đúng đề không bạn

25 tháng 8 2021

Thấy sai sai sao ấy