Câu hỏi của đoàn thế thịnh

Question

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

Nguyễn Lê Phước Thịnh · Accepted Answer

$A=\dfrac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\dfrac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\dfrac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}$

$=\dfrac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{\left(x^2+1-x\right)\left(x^2+1+x\right)}+\dfrac{\left(x-x^2+1\right)\left(x+x^2-1\right)}{\left(x^2+x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x-1\right)\left(x^2-x+1\right)}{\left(x^2-x-1\right)\left(x^2+x+1\right)}$

$=\dfrac{x^2+x-1}{x^2+x+1}+\dfrac{-x^2+x+1}{x^2+x+1}+\dfrac{x^2-x+1}{x^2+x+1}$

$=\dfrac{x^2+x-1-x^2+x+1+x^2-x+1}{x^2+x+1}$

$=\dfrac{x^2+x+1}{x^2+x+1}=1$

Câu trả lời:

$A=\dfrac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\dfrac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\dfrac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}$

$=\dfrac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{\left(x^2+1-x\right)\left(x^2+1+x\right)}+\dfrac{\left(x-x^2+1\right)\left(x+x^2-1\right)}{\left(x^2+x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x-1\right)\left(x^2-x+1\right)}{\left(x^2-x-1\right)\left(x^2+x+1\right)}$

$=\dfrac{x^2+x-1}{x^2+x+1}+\dfrac{-x^2+x+1}{x^2+x+1}+\dfrac{x^2-x+1}{x^2+x+1}$

$=\dfrac{x^2+x-1-x^2+x+1+x^2-x+1}{x^2+x+1}$

$=\dfrac{x^2+x+1}{x^2+x+1}=1$

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