\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)

=1/x-1/x+2014

\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)

a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

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

c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)

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

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)

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

a.

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

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

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

b.

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

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

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

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

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

c.

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

d.

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

\(=\dfrac{8+2\left(x-1\right)+x+3}{\left(x-1\right)\left(x+3\right)}=\dfrac{3x+9}{\left(x-1\right)\left(x+3\right)}\)

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

a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x

=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x

=8x.5(2x+1)(2x−1)(2

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

c) nè

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

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

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

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

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)\cdot2}{\left(x^2+1\right)\left(x-1\right)^2\left(x+1\right)}\)

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

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

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

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

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

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

b) \(=\dfrac{1}{x+2}+\dfrac{3}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\)

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

\(a,=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\\ b,=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{x+1}{x+3}\left(x\ne-1;x\ne-2;x\ne-3\right)\\ =\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)

\(a,\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)

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

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

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

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

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

\(b,\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)

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

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

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

\(=\dfrac{2x^2+8x+7}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)

\(=\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{\left(2x^2+8x+7\right).x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{2x^3+8x^2+7x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{2x^3+10x^2+15x+7}{\left(x+2\right)\left(x+3\right)^2}\)

\(\left(1-\frac{1}{x+1}\right)\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right)...\left(1-\frac{1}{x+2018}\right)\)

\(=\frac{x}{x+1}.\frac{x+1}{x+2}.\frac{x+2}{x+3}....\frac{x+2017}{x+2018}\)

\(=\frac{x}{x+2018}\)