Câu hỏi của Trần Thu Hiền

Question

HT.Phong (9A5) · Accepted Answer

a) $\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}$$=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(1-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x\left(1-x\right)}{x^2-9}$$=\dfrac{x^2+4x+3}{\left(x+3\right)\left(x-3\right)}-\dfrac{4x-3-x^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x-2x^2}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{x^2+4x+3-4x+3+x^2+2x-2x^2}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{6+2x}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{2\left(3+x\right)}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{2}{x-3}$b) $\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{\left(x+1\right)}+\dfrac{x+3}{1-x^2}$$=\dfrac{3x+1}{\left(1-x\right)^2}-\dfrac{1}{1+x}+\dfrac{x+3}{\left(1-x\right)\left(1+x\right)}$$=\dfrac{\left(3x+1\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)^2\left(1+x\right)}+\dfrac{\left(x+3\right)\left(1-x\right)}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{4x+3x^2+1}{\left(1-x\right)^2\left(1+x\right)}-\dfrac{1-2x+x^2}{\left(1-x\right)^2\left(1+x\right)}+\dfrac{-2x-x^2+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{4x+3x^2+1-1+2x-x^2-2x-x^2+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{x^2+4x+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{x+3}{\left(x-1\right)^2}$Câu trả lời: a) $\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}$$=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(1-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x\left(1-x\right)}{x^2-9}$$=\dfrac{x^2+4x+3}{\left(x+3\right)\left(x-3\right)}-\dfrac{4x-3-x^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x-2x^2}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{x^2+4x+3-4x+3+x^2+2x-2x^2}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{6+2x}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{2\left(3+x\right)}{\left(x+3\right)\left(x-3\right)}$$=\dfrac{2}{x-3}$b) $\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{\left(x+1\right)}+\dfrac{x+3}{1-x^2}$$=\dfrac{3x+1}{\left(1-x\right)^2}-\dfrac{1}{1+x}+\dfrac{x+3}{\left(1-x\right)\left(1+x\right)}$$=\dfrac{\left(3x+1\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)^2\left(1+x\right)}+\dfrac{\left(x+3\right)\left(1-x\right)}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{4x+3x^2+1}{\left(1-x\right)^2\left(1+x\right)}-\dfrac{1-2x+x^2}{\left(1-x\right)^2\left(1+x\right)}+\dfrac{-2x-x^2+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{4x+3x^2+1-1+2x-x^2-2x-x^2+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{x^2+4x+3}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(1-x\right)^2\left(1+x\right)}$$=\dfrac{x+3}{\left(x-1\right)^2}$

Nguyễn Lê Phước Thịnh · Answer

a: $=\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}$$=\dfrac{x^2+4x+3+x^2-4x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}$$=\dfrac{2}{x-3}$b: $=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}$$=\dfrac{3x^2+4x+1-x^2-2x+3-\left(x^2-2x+1\right)}{\left(x-1\right)^2\left(x+1\right)}$$=\dfrac{2x^2+2x+4-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x^2+4x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}$$=\dfrac{x+3}{\left(x-1\right)^2}$Câu trả lời: a: $=\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}$$=\dfrac{x^2+4x+3+x^2-4x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}$$=\dfrac{2}{x-3}$b: $=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}$$=\dfrac{3x^2+4x+1-x^2-2x+3-\left(x^2-2x+1\right)}{\left(x-1\right)^2\left(x+1\right)}$$=\dfrac{2x^2+2x+4-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x^2+4x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}$$=\dfrac{x+3}{\left(x-1\right)^2}$

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