Vì tử bằng mẫu

Ta có   \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{99}{100}}\)

\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)}\)

\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\)

\(\Rightarrow A=2\)

Ủa sao bạn ra được \(\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)  số 2 ở 200 đâu ra vậy ! và \(\frac{3}{2}\)nữa !

Đặt A= 200- (3+\(\frac{2}{3}+\frac{2}{4}+.....+\frac{2}{100}\))

         =\(197-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(\frac{197.2}{2}-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)

        =\(2.\left(\frac{196+1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

        =\(2\left(\frac{196}{2}+\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{100}\right)\)

         =\(2\left(98+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)

         =\(2\left(\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+.....+1-\frac{1}{100}\right)\)

          =\(2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.....+\frac{99}{100}\right)\)

Khi đó \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=2(đpcm)

Câu này trong đề kì thi Hà NỘi phải ko

ta có 200-(3+\(\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\)

=\(1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)

=\(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

thay \(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

ta có \(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\left(dpcm\right)\)

bvcdcvxcx

co cach tra loi khac day

Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)

=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=             \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=  \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)

\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)

cunasai

Đề là chứng minh điểu thức bằng 2 phải không bạn?

 \(\frac{200-\left(3+\frac{2}{3}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

\(=\frac{2.100-3-\frac{2}{3}-...-\frac{2}{100}}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

\(=\frac{\left(4-3\right)+\left(2-\frac{2}{3}\right)+\left(2-\frac{2}{4}\right)+...+\left(2-\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)

\(=\frac{1+\frac{4}{3}+\frac{6}{4}+...+\frac{198}{100}}{1+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)

\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)

\(=2\)

\(\)
 

\(\frac{200-\left(3+\frac{2}{3}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)

=\(\frac{2x100-3-\frac{2}{3}-...-\frac{2}{100}}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

=\(\frac{\left(4-3\right)+\left(2-\frac{2}{3}\right)+\left(2-\frac{2}{4}\right)+...+\left(2-\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

=\(\frac{1+\frac{4}{3}+\frac{6}{4}+...+\frac{198}{100}}{1+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)

=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)

=\(2\)

Trần Thùy Dung: (4-3) đâu ra vậy bạn?