Câu hỏi của Khoa

Question

rút gọn$\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}$$\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}$$\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}$

⭐Hannie⭐ · Accepted Answer

` @ \color{Red}{m}`` \color{lightblue}{Answer}`  $\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\
=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\
=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x}{x+1}$__$\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\
=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\
=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\
=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\
=\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\
=\dfrac{3x-2x+6}{2x\left(x+3\right)}\
=\dfrac{x+6}{2x\left(x+3\right)}$__$\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\
=\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1}{1+x}$Câu trả lời: ` @ \color{Red}{m}`` \color{lightblue}{Answer}`  $\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\
=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\
=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{x}{x+1}$__$\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\
=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\
=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\
=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\
=\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\
=\dfrac{3x-2x+6}{2x\left(x+3\right)}\
=\dfrac{x+6}{2x\left(x+3\right)}$__$\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\
=\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\
=\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\
=\dfrac{1}{1+x}$

YangSu · Answer

$\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)$$=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x}{x+1}$========================$\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)$$=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}$$=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}$$=\dfrac{3x-2x+6}{2x\left(x+3\right)}$$=\dfrac{x+6}{2x^2+6x}$==========================$\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)$$=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{1}{x+1}$Câu trả lời: $\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)$$=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x}{x+1}$========================$\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)$$=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}$$=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}$$=\dfrac{3x-2x+6}{2x\left(x+3\right)}$$=\dfrac{x+6}{2x^2+6x}$==========================$\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)$$=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}$$=\dfrac{1}{x+1}$

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)