a)A=(2x+1/2x-1-2x-1/2x+1) : 4x/10-5 b)B=(1/x^2+x-2-x/x+1) :(1/x+x-2)

1 bình luận về “a)A=(2x+1/2x-1-2x-1/2x+1) : 4x/10-5 b)B=(1/x^2+x-2-x/x+1) :(1/x+x-2)”

1. Giải đáp:$\begin{array}{l}a)A={\displaystyle \frac{10}{2x+1}}\\ b)B={\displaystyle \frac{1}{x+1}}\end{array}$

    

   Lời giải và giải thích chi tiết:

   $\begin{array}{l}Dkxd:x\ne {\displaystyle \frac{1}{2}};x\ne –{\displaystyle \frac{1}{2}};x\ne 0\\ A=\left({\displaystyle \frac{2x+1}{2x–1}}–{\displaystyle \frac{2x–1}{2x+1}}\right):{\displaystyle \frac{4x}{10x–5}}\\ ={\displaystyle \frac{{\left(2x+1\right)}^2–{\left(2x–1\right)}^2}{\left(2x–1\right)\left(2x+1\right)}}.{\displaystyle \frac{5.\left(2x–1\right)}{4x}}\\ ={\displaystyle \frac{4x^2+4x+1–4x^2+4x–1}{2x+1}}.{\displaystyle \frac{5}{4x}}\\ ={\displaystyle \frac{8x}{2x+1}}.{\displaystyle \frac{5}{4x}}\\ ={\displaystyle \frac{10}{2x+1}}\\ b)Dkxd:x\ne 0;x\ne –1\\ B=\left({\displaystyle \frac{1}{x^2+x}}–{\displaystyle \frac{2–x}{x+1}}\right):\left({\displaystyle \frac{1}{x}}+x–2\right)\\ =\left({\displaystyle \frac{1}{x\left(x+1\right)}}–{\displaystyle \frac{2–x}{x+1}}\right):{\displaystyle \frac{1+x\left(x–2\right)}{x}}\\ ={\displaystyle \frac{1–x.\left(2–x\right)}{x\left(x+1\right)}}.{\displaystyle \frac{x}{1+x^2–2x}}\\ ={\displaystyle \frac{1–2x+x^2}{x+1}}.{\displaystyle \frac{1}{x^2–2x+1}}\\ ={\displaystyle \frac{1}{x+1}}\end{array}$

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