Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

2.

1/2004

sudy well

study well

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

minh ko biet nua moi lop 4 thoi

hình như là sai đề

Áp dụng tc của dãy tỉ số bằng nhau ta có

\(\frac{x_1-1}{10}=.....=\frac{x_{10}-10}{1}=\frac{\left(x_1+x_2+....+x_{10}\right)-\left(1+2+3+...+10\right)}{1+2+3+...+10}\)

\(=\frac{45}{55}=\frac{9}{11}\)

Giải ra ta được

\(x_1=\frac{101}{11}\)

\(x_2=\frac{103}{11}\)

........

\(x_{10}=\frac{119}{11}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:: \(\frac{x_1-1}{10}=\frac{x_2-2}{9}=...=\frac{x_9-9}{2}=\frac{x_{10}-10}{1}=\frac{x_1-1+x_2-2+...+x_9-9+x_{10}-10}{10+9+...+2+1}\)

\(=\frac{\left(x_1+x_2+..+x_9+x_{10}\right)-\left(1+2+...+9+10\right)}{10+9+...+2+1}=\frac{100-55}{55}=\frac{9}{11}\)

=> \(\frac{x_1-1}{10}=\frac{9}{11}\Leftrightarrow x_1-1=\frac{10\cdot9}{11}\Leftrightarrow x_1=\frac{90}{11}+1=\frac{101}{11}\)

     \(\frac{x_2-2}{9}=\frac{9}{11}\Leftrightarrow x_2-2=\frac{9\cdot9}{11}\Leftrightarrow x_2=\frac{81}{11}+2=\frac{103}{11}\)

Tương tự ta cũng có:

     \(x_3=\frac{105}{11}\)

    \(x_4=\frac{107}{11}\)

    \(x_5=\frac{109}{11}\)

     \(x_6=\frac{111}{11}\)

    \(x_7=\frac{113}{11}\)

    \(x_8=\frac{115}{11}\)

    \(x_9=\frac{117}{11}\)

   \(x_{10}=\frac{119}{11}\)

d) \(\frac{x}{-9}=\left(\frac{2}{6}\right)^2\)

\(\Rightarrow\frac{x}{-9}=\frac{2}{6}.\frac{2}{6}\)

\(\Rightarrow\frac{x}{-9}=\frac{4}{36}\)

\(\Rightarrow\frac{x}{-9}=\frac{1}{9}\)

\(\Rightarrow\frac{-x}{9}=\frac{1}{9}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=1\)

e) \(\frac{a}{b}+\frac{3}{6}=0\)

\(\Rightarrow\frac{a}{b}=0-\frac{3}{6}\)

\(\Rightarrow\frac{a}{b}=0-\frac{1}{2}\)

\(\Rightarrow\frac{a}{b}=\frac{-1}{2}\)

\(\Rightarrow a=-1;b=2\)

x+2/327 + x+3/326 + x+4/325 + x+5/324 + x+349/5 = 0

=> x+2/327 + 1 + x+3/326 + 1 + x+4/325 + 1 + x+5/324 + 1 - x+349/5 - 4 = 0

=> x+329/327 + x+329/326 + x+329/325 + x+329/324 + x+329/5 = 0

=> (x+329).(1/327 + 1/326 + 1/325 + 1/324 + 1/5) = 0

Dễ thấy: 1/327 + 1/326 + 1/325 + 1/324 + 1/5 > 0

=> x + 329 = 0

=> x = -329