a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)

\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)

\(\Leftrightarrow8x=-\frac{5}{4}\)

\(\Leftrightarrow x=-\frac{5}{32}\)

c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)

\(\Leftrightarrow x+1=2003\)

\(\Leftrightarrow x=2002\)

Ta có :

\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)

\(\Rightarrow A=2009\)

1)\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2008+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{1+\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{\dfrac{2009}{2009}+\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}}\)

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}{2009\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}\)

\(\dfrac{A}{B}=\dfrac{1}{2009}\)

2) \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)

Đặt \(C=1+2+2^2+...+2^{2007}+2^{2008}\)

\(\Rightarrow2C=2+2^2+2^3+...+2^{2008}+2^{2009}\)

\(\Rightarrow2C-C=2^{2009}-1\)

\(\Rightarrow C=2^{2009}-1\)

\(\Rightarrow B=\dfrac{2^{2009}-1}{1-2^{2009}}=\dfrac{-1\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

Giải:

B=1+2+2²+2³+...+2²⁰⁰⁸/1-2²⁰⁰⁹

Ta gọi phần tử là A, ta có:

A=1+2+2²+2³+...+2²⁰⁰⁸

2A=2+2²+2³+2⁴+...+2²⁰⁰⁹

2A-A=(2+2²+2³+2⁴+...+2²⁰⁰⁹)-(1+2+2²+2³+...+2²⁰⁰⁸)

A=2²⁰⁰⁹-1

Vậy B=2²⁰⁰⁹-1/1-2²⁰⁰⁹

Chúc bạn học tốt!

a)

`2/3+5/2-3/4`

`=10/4-3/4+2/3`

`=7/4+2/3`

`=21/12+8/12`

`=29/12`

b)

`2/5xx1/2:1/3`

`=2/10xx3/1`

`=6/10=3/5`

c)

`2/9:2/9xx1/3`

`=2/9xx9/2xx1/3`

`=1xx1/3`

`=1/3`

a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)

= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)

= \(\dfrac{38-9}{12}\)

= \(\dfrac{29}{12}\)

b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)

= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)

= \(\dfrac{3}{5}\)

c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)

= 1 x \(\dfrac{1}{3}\)

= \(\dfrac{1}{3}\)

\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)

Bài này dễ nên nãy không có định làm, nhưng mà nghĩ lại thôi làm vậy:Đ

a/ \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\)

Ta có: \(\dfrac{-4}{y}=\dfrac{2}{4}\Rightarrow y=\dfrac{-4.4}{2}=-8\)

\(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{-4}{-8}=\dfrac{1}{2}\Rightarrow x=\dfrac{-2.1}{2}=-1\)

b/\(\dfrac{2}{x}=\dfrac{y}{-3}\Rightarrow xy=-3.2=-6\)

\(\Rightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta có bảng giá trị của x, y như sau:

c/ \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\Rightarrow\left(x+1\right)^2=2.8=16\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=4^2\\\left(x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

c,(x+1).(x+1)=8.2

(x+1)²=16

⇒x+1=4

⇒x=3

a: =11/2*4*5/3

=22*5/3

=110/3

b: =30/12-3/12+20/12

=47/12

c: =28/15+5

=28/15+75/15

=103/15