Câu hỏi của Kwalla

Question

Thực hiện phép tính 1/x+2 + 5/2x2+3x-2-3x2/x3+11  +  1/x2-x+1  +1/x+11/1-x  +1/1+x  +2/1+x2  +4/1+x4

Toru · Accepted Answer

a) $\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}$$=\dfrac{1}{x+2}+\dfrac{5}{2x^2+4x-x-2}$$=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{2x\left(x+2\right)-\left(x+2\right)}$$=\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2}{2x-1}$$---$b) $\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}$ (sửa đề)$=\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2x+2}{x^2-x+1}$$---$c) $\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{1+x+1-x}{1^2-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}$$=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}$$=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}$$=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}$$=\dfrac{4+4x^4+4-4x^4}{1-x^8}$$=\dfrac{8}{1-x^8}$#$Toru$Câu trả lời: a) $\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}$$=\dfrac{1}{x+2}+\dfrac{5}{2x^2+4x-x-2}$$=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{2x\left(x+2\right)-\left(x+2\right)}$$=\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}$$=\dfrac{2}{2x-1}$$---$b) $\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}$ (sửa đề)$=\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2x+2}{x^2-x+1}$$---$c) $\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{1+x+1-x}{1^2-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}$$=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}$$=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}$$=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}$$=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}$$=\dfrac{4+4x^4+4-4x^4}{1-x^8}$$=\dfrac{8}{1-x^8}$#$Toru$

⭐Hannie⭐ · Answer

$\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\ =\dfrac{1}{x+2}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2}{2x-1}$__`x^3+1` chứ cậu nhỉ?$\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x-1\right)\left(x^2-x+1\right)}\ =\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2\left(x-1\right)}{x^2-x+1}$__ Câu trả lời: $\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\ =\dfrac{1}{x+2}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}\ =\dfrac{2}{2x-1}$__`x^3+1` chứ cậu nhỉ?$\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\ =\dfrac{-3x^2}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x-1\right)\left(x^2-x+1\right)}\ =\dfrac{-3x^2+x+1+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2\left(x^2-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$$=\dfrac{-2\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\ =\dfrac{-2\left(x-1\right)}{x^2-x+1}$__

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