Bài dễ ẹt

    2003 x2005 +200

    2004x2004+199

= 1

tk nhé

Tớ nghĩ là .................................................botay.kom

\(\frac{2005\times2004-1}{2003\times2005+2004}\)

=\(\frac{2005\times2004-1}{2003\times2005+\left(2005-1\right)}\)

=\(\frac{2005\times2004-1}{2003\times2005+2005-1}\)

=\(\frac{2005\times2004-1}{\left(2003+1\right)\times2005-1}\)

=\(\frac{2005\times2004-1}{2004\times2005-1}\)

=1

(2004-1)(2004+1)+200/2004.2004+199

=2004^2-1+200/2004^2+199

=2004^2+199/2004^2+199

=1

\(\frac{2003.2005+200}{2004.2004+199}\)

= \(\frac{2003.\left(2004+1\right)+200}{\left(2003+1\right).2004+199}\)

= \(\frac{2003.2004+2003+200}{2003.2004+2004+199}\)

= \(\frac{2003.2004+2203}{2003.2004+2203}\)

= \(1\)

                 ~~~~Hok tốt~~~~

\(\frac{\left(16-8:5\right)x177}{199x2001}\frac{\left(16-16\right)x177}{199x2001}=\frac{0x177}{199x2001}=\frac{0}{199x2001}=0\)

Theo cách mk học sẽ suy ra lun

=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2001-1/2003+1/2003-1/2005

=1-1/2005

=2004/2005

1995(1996+1)-1/1996.1995+1994

1995.1996+1995-1/1996.1995+1994

1995.1996+1994/1996.1995+1994=1

\(\frac{1995.1997-1}{1996.1995+199}\)= \(\frac{1995.\left(1996+1\right)-1}{1996.1995+199}\)

                         = \(\frac{1995.1996+1995.1-1}{1996.1995+199}\)

                         = \(\frac{1995.1996+1994}{1996.1995+199}\)

làm đến đây thì mình chịu

= 2004 - 1 / 2003  + 2004

= 2003 / 2007

mình ko biết có đúng ko nữa         

Cho mk sửa lại đáp án là \(\frac{100}{201}\)nha bn

Ta có: \(N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)

\(\Rightarrow2N=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\)

\(\Rightarrow2N=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)

\(\Rightarrow2N=\frac{1}{1}-\frac{1}{201}\)

\(\Rightarrow2N=\frac{200}{201}\)

\(\Rightarrow N=\frac{200}{201}:2=\frac{100}{101}\)

tk cho mk nha bn

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