Mình xin phép sửa đề 1 trust ạ :>

Xác định các số a,b,c sao cho \(\frac{1}{\left(x^2+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\)     

Điều kiện x khác 1 :vv

\(pt\Leftrightarrow\frac{1}{\left(x^2+1\right)\left(x-1\right)}=\frac{\left(ax+b\right)\left(x-1\right)}{\left(x^2+1\right)\left(x-1\right)}+\frac{c\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)

\(\Leftrightarrow1=ax^2-ax+bx-b+cx^2+c\)

\(\Leftrightarrow\left(a+c\right)x^2+\left(b-a\right)x+\left(c-b-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a+c=0\\b-a=0\\c-b-1=0\end{cases}\Leftrightarrow}a=-\frac{1}{2};b=-\frac{1}{2};c=\frac{1}{2}\)

Vậy .....

B = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x........x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

B = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x.........x\frac{2002}{2003}x\frac{2003}{2004}\)

=> B = \(\frac{1}{2004}\)

\(B=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)

\(B=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2002}{2003}\times\frac{2003}{2004}\)

\(B=\frac{1\times2\times3\times...\times2002\times2003}{2\times3\times4\times...\times2003\times2004}\)

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

Ta có:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2002}{2003}.\frac{2003}{2004}\)

\(B=\frac{1.2.3.4....2002.2003}{2.3.4.5....2003.2004}\)

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

#)Giải :

\(\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{97}\right)\left(1-\frac{2}{99}\right)\)

\(=\frac{3}{5}\times\frac{5}{7}\times\frac{7}{9}\times...\times\frac{95}{97}\times\frac{97}{99}\)

\(=\frac{3}{99}=\frac{1}{33}\)( loại các cặp số giống nhau ở tử và mẫu của các phân số )

\(\left[1-\frac{2}{5}\right]\times\left[1-\frac{2}{7}\right]\times\left[1-\frac{2}{9}\right]\times\left[1-\frac{2}{11}\right]\times...\times\left[1-\frac{2}{97}\right]\times\left[1-\frac{2}{99}\right]\)

\(=\frac{3}{5}\times\frac{5}{7}\times\frac{7}{9}\times\frac{9}{11}\times...\times\frac{95}{97}\times\frac{97}{99}\)

\(=\frac{3\times5\times7\times9\times...\times97}{5\times7\times9\times11\times...\times97\times99}=\frac{3}{99}=\frac{1}{33}\)

\(\left[1-\frac{2}{5}\right]\times\left[1-\frac{2}{7}\right]\times\left[1-\frac{2}{9}\right]\times...\times\left[1-\frac{2}{97}\right]\times\left[1-\frac{2}{99}\right]\)

=\(\frac{3}{5}\times\frac{5}{7}\times\frac{7}{9}\times...\times\frac{95}{97}\times\frac{97}{99}\)

=\(\frac{3\times5\times7\times...95\times97}{5\times7\times9\times...\times97\times99}\)

=\(\frac{3}{99}\)

=\(\frac{1}{33}\)

tim ban voi ai do rui tra loi cho tui ko bt

\(\left(1-\frac{1}{x^2}\right)\left(1-\frac{1}{y^2}\right)=1+\frac{2}{xy}\)

<=>  \(1-\frac{1}{x^2}-\frac{1}{y^2}+\frac{1}{x^2y^2}=1+\frac{2}{xy}\)

<=>   \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}-\frac{1}{x^2y^2}=0\)

<=>   \(\left(\frac{1}{x}+\frac{1}{y}\right)^2-\frac{1}{x^2y^2}=0\)

<=>  \(\left(\frac{x+y}{xy}\right)^2-\frac{1}{x^2y^2}=0\)

<=>  \(\frac{1}{x^2y^2}-\frac{1}{x^2y^2}=0\)   luôn đúng

=>  đpcm

\(\dfrac{3}{5}\) x \(\dfrac{1}{7}\) x 1\(\dfrac{1}{2}\)

= \(\dfrac{3}{35}\) x \(\dfrac{3}{2}\)

= \(\dfrac{9}{70}\)

\(\dfrac{3}{5}\times\dfrac{4}{7}\times1\dfrac{1}{2}\)

\(=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{3}{2}\)

\(=\dfrac{3\times4\times3}{5\times7\times2}\)

\(=\dfrac{3\times2\times3}{5\times7}\)

\(=\dfrac{18}{35}\)