2008 nhé chúc năm mới vui vẻ

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(x+1=2009\)

\(x=2008\)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dots+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2008}{2009}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2009-1\)

\(\Rightarrow x=2008\)

\(\dfrac{3}{7}\cdot\dfrac{5}{4}-\dfrac{5}{7}\cdot\dfrac{1}{3}-\dfrac{1}{7}\cdot\dfrac{5}{12}\)

\(=\dfrac{15}{28}-\dfrac{5}{21}-\dfrac{5}{84}\)

\(=\dfrac{25}{84}-\dfrac{5}{84}\)

\(=\dfrac{5}{21}\)

5/21  chúc năm mới vui vẻ

\(\dfrac{4}{7}x-x=-\dfrac{9}{14}\)

\(\left(\dfrac{4}{7}-1\right)x=-\dfrac{9}{14}\)

\(-\dfrac{3}{7}x=-\dfrac{9}{14}\)

\(x=-\dfrac{9}{14}:\dfrac{-3}{7}\)

\(x=\dfrac{3}{2}\)

\(\dfrac{4}{7}x-x=-\dfrac{9}{14}\)

\(\Rightarrow x\cdot\left(\dfrac{4}{7}-1\right)=-\dfrac{9}{14}\)

\(\Rightarrow x\cdot\dfrac{-3}{7}=\dfrac{-9}{14}\)

\(\Rightarrow x=\dfrac{-9}{14}:\dfrac{-3}{7}\)

\(\Rightarrow x=\dfrac{3}{2}\)

Bạn chỉnh lại môn học nhé.

Bài 2:

d; \(\dfrac{3}{5}\) : (-5) = \(\dfrac{3}{5}\) x \(\dfrac{1}{\left(-5\right)}\) = \(\dfrac{3}{-25}\) = \(\dfrac{-3}{25}\)

e; \(\dfrac{-4}{15}\): 2 = \(\dfrac{-4}{15}\) x \(\dfrac{1}{2}\) = \(\dfrac{-4}{30}\) 

f; 24 : \(\dfrac{-6}{7}\) = 24 x \(\dfrac{7}{-6}\) = \(\dfrac{28}{-1}\) = -28

Bài 3:

b; \(\dfrac{-4}{5}\) : \(\dfrac{2}{7}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-4}{5}.\dfrac{7}{2}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-14}{5}\) + \(\dfrac{4}{7}\)

= \(\dfrac{-98}{35}\) + \(\dfrac{20}{35}\)

= \(\dfrac{-78}{35}\)

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

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

\(\dfrac{1}{10}x=-\dfrac{2}{3}\)

\(x=-\dfrac{20}{3}\)

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

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

\(\Rightarrow x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)

\(\Rightarrow x=-\dfrac{2}{3}:\dfrac{11}{10}\)

\(\Rightarrow x=-\dfrac{20}{33}\)

Lời giải:
Gọi $d=ƯCLN(3n+1, n(3n+2))$

$\Rightarrow 3n+1\vdots d; n(3n+2)\vdots d$

$\Rightarrow n(3n+1)+n\vdots d$

Mà $3n+1\vdots d\Rightarrow n\vdots d$

$\Rightarrow 3n+1-3n\vdots d$

$\Rightarrow 1\vdots d\Rightarrow d=1$

Vậy ps $\frac{3n+1}{n(3n+2)}$ là phân số tối giản.

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

\(\Rightarrow\dfrac{3}{4}:x=-\dfrac{1}{2}-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{3}{4}:x=-\dfrac{9}{10}\)

\(\Rightarrow x=\dfrac{3}{4}:-\dfrac{9}{10}\)

\(\Rightarrow x=-\dfrac{5}{6}\)

Vậy; ...

\(\dfrac{5355}{918}=\dfrac{5355:153}{918:153}=\dfrac{35}{6}\)

35/6

\(\dfrac{1}{2}a^2+1=2\) 

\(\Rightarrow\dfrac{1}{2}a^2=2-1\)

\(\Rightarrow\dfrac{1}{2}a^2=1\)

\(\Rightarrow a^2=2\)

\(\Rightarrow a^2=\left(\sqrt{2}\right)^2\)

\(\Rightarrow a\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

\(\dfrac{1}{2a^2+1}=2\)

\(2a^2+1=1\)

\(2a^2=0\)

\(a^2=0\)

\(a=0\)