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\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

4 tháng 12 2018

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

4 tháng 12 2018

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

29 tháng 11 2017

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

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

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

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

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

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

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

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

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

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

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

2) \(\dfrac{5}{x+1}-\dfrac{10}{x-x^2-1}-\dfrac{15}{x^3+1}\)

\(=\dfrac{5}{x+1}-\dfrac{10}{-\left(x^2-x+1\right)}-\dfrac{15}{x^3+1}\)

\(=\dfrac{5}{x+1}+\dfrac{10}{\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

\(=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{10\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

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

29 tháng 11 2017

3) \(\dfrac{2}{2x+1}-\dfrac{1}{2x-1}-\dfrac{2}{1-4x^2}\)

\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{4x^2-1}\)

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

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

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

\(=\dfrac{4x-2-2x-1+2}{\left(2x-1\right)\left(2x+1\right)}\)

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

\(=\dfrac{1}{2x+1}\)

4) \(\dfrac{3x^2+5x+14}{x^3+1}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\)

\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{\left(3x^2+5x+14\right)+\left(x-1\right)\left(x+1\right)-4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

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

24 tháng 6 2017

Phân thức đại số

Phân thức đại số

28 tháng 6 2017

Phép cộng các phân thức đại số

Phép cộng các phân thức đại số

4 tháng 12 2018

a.\(\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) = \(\frac{5x+1-1+3x-2x^2+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\) =\(\frac{10x+2}{x^3-1}\)

b.\(\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{x^3+1}\)( đến đây dễ r đúng ko)

4 tháng 12 2018

Dễ thế mak ko biết làm, đồ ngu như óc chó

4 tháng 12 2018

Cứ làm như og giỏi lắm ế

Người ta thấy khó thì người ta hỏi

Có chết đâu

26 tháng 11 2017

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

\(b,\dfrac{3x+5}{x^2-5x}+\dfrac{x-25}{5x-25}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)

\(=\dfrac{5\left(3x+5\right)}{5x\left(x-5\right)}+\dfrac{\left(x-25\right)x}{5x\left(x-5\right)}\)

\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}\)

\(=\dfrac{x^2-10x+25}{5x\left(x-5\right)}\)

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

\(c,\left(\dfrac{2}{x-1}-\dfrac{2}{x+1}\right).\dfrac{x^2+2x+1}{4}\)

\(=\left(\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(x+1\right)^2}{4}\)

\(=\dfrac{2x+2-2x+2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)^2}{4}\)

\(=\dfrac{4}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)^2}{4}\)

\(=\dfrac{x+1}{x-1}\)

29 tháng 6 2017

Phép chia các phân thức đại số

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

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

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

 

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)