d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(\dfrac{1}{3}-\dfrac{1}{101}\)

= \(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=\dfrac{10}{39}+\dfrac{2}{39}\)

\(=\dfrac{4}{13}\)

gọi biểu thức đó là A

A=\(1.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)+\dfrac{2}{39}\)

A= \(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)+\dfrac{2}{39}=\dfrac{4}{13}\)

mk nhanh nhất nha bạn

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)

Chúc bạn học tốt!!!

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

= \(\dfrac{1}{3}-\dfrac{1}{15}\)

= \(\dfrac{4}{15}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=1/2*10/39

=5/39

2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)

I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)

\(\Leftrightarrow3n+6-2⋮n+2\)

\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{-1;-3;0;-4\right\}\)

II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)

có nhất thiết phải hỏi mấy bài thế này ko ?

125%.$(\frac{-1}{2}{)}^2$:$(1\frac{5}{16}-1,5)+{2008}^0$

$\frac{\mathrm{=1.25x0,25:(1,3125-1,5)+1}}{}$

$\frac{\mathrm{=0,3125:(-0,1875)+1}}{}$

$\frac{\mathrm{=-5/3+1}}{}$

=-2/3

vậy.......

$\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}$

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)

câu này viết sai đề nhé!!!!

Bài 2. Tìm x, biết

x:$3\frac{1}{15}=1\frac{1}{12}$

$\frac{\mathrm{\backslash (\backslash Rightarrow\; x=1\backslash dfrac\{1\}\{12\}\backslash cdot3\backslash dfrac\{1\}\{15\}=\backslash dfrac\{299\}\{90\}\backslash )}}{}$

vậy

$\frac{3}{4}.x=27$

$\mathrm{\backslash (\backslash Rightarrow\; x=27:\backslash dfrac\{3\}\{4\}=36\backslash )}$

vậy.....

$x.\frac{1}{4}=\frac{5}{8}.\frac{2}{3}$

\(\Rightarrow x=\dfrac{5}{8}\cdot\dfrac{2}{3}:\dfrac{1}{4}=\dfrac{5}{3}\)

vậy........

$\left(x+\frac{1}{4}\right):\frac{3}{7}=\frac{1}{6}$

$\frac{\mathrm{\backslash (x:\backslash dfrac\{3\}\{7\}+\backslash dfrac\{1\}\{4\}:\backslash dfrac\{3\}\{7\}=\backslash dfrac\{1\}\{6\}\backslash )}}{}$

\(x:\dfrac{3}{7}=\dfrac{1}{6}-\dfrac{1}{4}:\dfrac{3}{7}\)

\(x:\dfrac{3}{7}=-\dfrac{5}{12}\)

\(x=-\dfrac{5}{12}\cdot\dfrac{3}{7}=-\dfrac{5}{28}\)

vậy............

$\frac{3}{4}-2.|2x-\frac{2}{3}|=2$

$\mathrm{\backslash (\backslash dfrac\{3\}\{4\}-4\backslash left|x-\backslash dfrac\{1\}\{3\}\backslash right|=2\backslash )}$

\(4\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}-2\)

\(4\left|x-\dfrac{1}{3}\right|=-1,25\)

\(\left|x-\dfrac{1}{3}\right|=-1,25:4=-0,3125\)(loại)

vậy.............

$\frac{2}{3}$$x-\frac{1}{2}x=\frac{5}{2}$