Câu hỏi của Lê Mai Tuyết Hoa

Question

a, $\dfrac{1}{a+1}+\dfrac{2}{1-a}+\dfrac{5a-1}{a^2-1}$         b, $\dfrac{1}{9x-18}+\dfrac{16-7x}{72-18x}+\dfrac{5}{12x-24}$

hưng phúc · Accepted Answer

$a.\dfrac{1}{a+1}+\dfrac{2}{1-a}+\dfrac{5a-1}{a^2-1}$$=\dfrac{1}{a+1}-\dfrac{2}{a-1}+\dfrac{5a-1}{a^2-1}$$=\dfrac{a-1}{\left(a+1\right)\left(a-1\right)}-\dfrac{2\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}+\dfrac{5a-1}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{a-1-2a-2+5a-1}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{4a-4}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{4}{a+1}$Câu trả lời: $a.\dfrac{1}{a+1}+\dfrac{2}{1-a}+\dfrac{5a-1}{a^2-1}$$=\dfrac{1}{a+1}-\dfrac{2}{a-1}+\dfrac{5a-1}{a^2-1}$$=\dfrac{a-1}{\left(a+1\right)\left(a-1\right)}-\dfrac{2\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}+\dfrac{5a-1}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{a-1-2a-2+5a-1}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{4a-4}{\left(a+1\right)\left(a-1\right)}$$=\dfrac{4}{a+1}$

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