\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-...-\frac{98}{100}+\frac{99}{100}-\frac{100}{100}\)

\(=\frac{1-2+3-...-98+99-100}{100}\)

\(=\frac{\left[\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+\left(99-100\right)\right]}{100}\)

\(=\frac{-1-1-1-...-1}{100}=\frac{-1.50}{100}=\frac{-50}{100}=\frac{-1}{2}\)

Vậy S=\(\frac{-1}{2}\)

\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-\frac{4}{100}+\frac{5}{100}-...-\frac{98}{100}+\frac{99}{100}\)

\(S=\frac{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+....+\left(97-98\right)+\left(99-100\right)}{100}\)

\(S=\frac{-1+\left(-1\right)+\left(-1\right)+.....+\left(-1\right)+\left(-1\right)}{100}\)

Từ 1 đến 100 có 100 số số hạng => Có 50 cặp => có 50 số (-1)

=> \(S=\frac{50\cdot\left(-1\right)}{100}=\frac{-50}{100}=\frac{-1}{20}\)

tseêre567889933333

\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{100}{2^{101}}\)

\(A-\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

\(\frac{1}{2}A=\left(1-\frac{1}{2^{101}}\right)\div\frac{1}{2}-\frac{100}{2^{101}}\)

\(=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)

\(\Rightarrow A=\frac{\left(2^{101}-1\right)}{2^{99}}-\frac{100}{2^{100}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)

=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

=>\(A=2A-A=2+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

\(A=2+\frac{1}{2^{98}}\)

Vậy: \(A=2+\frac{1}{2^{98}}\)

Gọi \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2B-B=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow B=2-\frac{1}{2^{100}}\)

\(\Rightarrow A=2\)

Vậy A = 2

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}+\frac{1}{2^{101}}\)

Suy ra \(2.A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

Nên \(2.A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}+\frac{1}{2^{101}}\right)\)

Khi đó \(A=2-\frac{1}{2^{101}}\)

Vậy \(A=2-\frac{1}{2^{101}}\)

Sửa đề:

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

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

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

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

haha!dungs rois!

sai đề bài kìa

Tách 100 thành 100 số 1

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

=\(0+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)=MS

=> Phân số trên=1

mot ban co 309 cai keo hoi 23 ban bao nhieu cai keo/

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

đề sai r bạn