\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{13\cdot15}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}=\dfrac{14}{15}\)

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

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

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

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

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

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

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

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

a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)

\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)

\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)

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

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

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

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

Câu b/ quên làm ạ :> Bù nè

b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)

\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)

\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)

\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)

\(=12x^2+26x-10-12x^2+11x-2\)

\(=37x-12\)

thanks ah

\(\left(5x^4-x^3+3x^2\right)\)

5x⁴-x³+3x²

a,     (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))

=  (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\))  \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\)  - \(\dfrac{21}{60}\))

= - \(\dfrac{3}{80}\)  \(\times\) (- \(\dfrac{2}{3}\))

= \(\dfrac{1}{40}\) 

b, (-1)³ + (- \(\dfrac{2}{3}\))² : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)

=  -1³ +   \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1

= 0

c, 15 \(\times\) (- \(\dfrac{1}{5}\))² + \(\dfrac{1}{5}\) - 2 \(\times\) (- \(\dfrac{1}{2}\))³ - \(\dfrac{1}{2}\)

= 15 \(\times\) \(\dfrac{1}{25}\) + \(\dfrac{1}{5}\) - 2 \(\times\) (- \(\dfrac{1}{8}\)) - \(\dfrac{1}{2}\)

=  \(\dfrac{3}{5}\)  + \(\dfrac{1}{5}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{2}\)

= \(\dfrac{4}{5}\) - \(\dfrac{1}{4}\)

= \(\dfrac{11}{20}\)

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

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)

\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)

\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)

\(=5-\dfrac{4}{2}\)

\(=5-2\)

\(=3\)

b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)

\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)

\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)

\(=0+0+0+2022\)

\(=2022\)

2) \(0,7^2\cdot x=0,49^2\)

\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)

\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)

\(\Rightarrow x=\left(0,7\right)^2\)

\(\Rightarrow x=0,49\)

b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)

\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)

\(\Rightarrow x=\left(-0,5\right)^5\)

\(\Rightarrow x=-\dfrac{1}{32}\)

2:

a: =>x*0,49=0,49^2

=>x=0,49

b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5