Câu hỏi của Nguyễn Hải Anh

a: \(A=\dfrac{-2x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x^2+x+1}\)\(=\dfrac{-2x^2+x^2+x+1+x^2-x}{x\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x\left(x-1\right)\left(x^2+x+1\right)}\)Khi x=10 thì \(A=\dfrac{1}{10\left(10-1\right)\left(10^2+10+1\right)}=\dfrac{1}{9990}\)b: \(B=\dfrac{-x^4+\left(x-1\right)\left(x^3+x^2+x+1\right)}{x-1}+1\)\(=\dfrac{-x^4+x^4-1}{x-1}+1=\dfrac{-1}{x-1}+1=\dfrac{-1}{-100}+1=\dfrac{1}{100}+1=\dfrac{101}{100}\)Câu trả lời: a: \(A=\dfrac{-2x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x^2+x+1}\)\(=\dfrac{-2x^2+x^2+x+1+x^2-x}{x\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x\left(x-1\right)\left(x^2+x+1\right)}\)Khi x=10 thì \(A=\dfrac{1}{10\left(10-1\right)\left(10^2+10+1\right)}=\dfrac{1}{9990}\)b: \(B=\dfrac{-x^4+\left(x-1\right)\left(x^3+x^2+x+1\right)}{x-1}+1\)\(=\dfrac{-x^4+x^4-1}{x-1}+1=\dfrac{-1}{x-1}+1=\dfrac{-1}{-100}+1=\dfrac{1}{100}+1=\dfrac{101}{100}\)