Bài 5: Phép cộng các phân thức đại số

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

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x-4}{x-2}\)

Làm chi tiết theo yêu cầu của bạn :

\(ĐKXĐ\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x-4}{x-2}\)

p/s : bonus nếu bạn không hiểu ptich đa thức thành nhân tử =))

x² + x - 6 = x² + 3x - 2x - 6 = x(x+3) - 2(x+3) = (x-2)(x+3)

x² - x - 12 = x² + 3x - 4x - 12 = x(x+3) - 4(x+3) = (x-4)(x+3)

a, \(\dfrac{x^2-49}{x-7}\) + x - 2 = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x - 2 = x + 7 + x - 2 = 2x + 5

b, \(\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right)\) . \(\dfrac{x^2+6x}{2x-6}\) 

= \(\left(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\left(\dfrac{6\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right)\) . \(\dfrac{x\left(x+6\right)}{2x-6}\)

= \(\dfrac{6}{x-6}\)

1. = \(\dfrac{\left(x-7\right)\left(x+7\right)}{x-7}\) + x-2

    = x+7 +x-2

    = 2x-5

2.  = (\(\dfrac{x}{\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{x-6}{x\left(x+6\right)}\) ) \(^{\dfrac{x^2+6x}{2x-6}}\)

     = ( \(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}\) - \(\dfrac{\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\) ) \(\dfrac{x^2+6x}{2x-6}\) 

     = \(\dfrac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\)  . \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\) .  \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\) . \(\dfrac{x^2+6x}{2x-6}\)

     = \(\dfrac{12\left(x-3\right)x\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)2\left(x-3\right)}\)

     = \(\dfrac{6}{x-6}\)

Chúc bạn học tốt!

\(x^2+6x+9=0\\ \Leftrightarrow x^2+2.3.x+3^2=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

Vậy \(x=-3\)

\(\dfrac{x+1}{x-3}\) + \(\dfrac{x-1}{x+3}\) - \(\dfrac{2x-2x^2}{9-x^2}\)

= \(\dfrac{x+1}{x-3}\)+ \(\dfrac{x-1}{x+3}\) + \(\dfrac{2x-2x^2}{x^2-9}\)

= \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\) + \(\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\) + \(\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{x^2+3x+x+3+\left(x^2-3x-x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{x^2+4x+3+x^2-3x-x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

    \(\dfrac{7}{x}+\dfrac{3}{x-3}-\dfrac{4x-3}{x^2-3x}\)

\(=\dfrac{7\left(x-3\right)}{x\left(x-3\right)}+\dfrac{3x}{x\left(x-3\right)}-\dfrac{4x-3}{x\left(x-3\right)}\)

\(=\dfrac{7x-21+3x-4x+3}{x\left(x-3\right)}\)

\(=\dfrac{6x-18}{x\left(x-3\right)}\)

\(=\dfrac{6\left(x-3\right)}{x\left(x-3\right)}\)

\(=\dfrac{6}{x}\)

Bạn viết không đủ đề! Mình có thể xoá đấy ạ!

À không chắc nhé!