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b)ta đặt A: \(A=\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\)
\(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+..+\left(\frac{98}{2}+1\right)+\left(\frac{99}{1}-98\right)\)
\(A=\frac{100}{99}+\frac{100}{98}+..+\frac{100}{2}+\frac{100}{100}\)
\(A=100\cdot\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}\right)\)
1/3 = 1515/4545
2:2/3 = 3
1 va 1/3 = 1/6 . 8
5/6 = 2 va 1/3-1 va 1/2 ( cau nnay viet de bi sai)
7/12 = 1/3+1/4
1/5 = 1/3 . 3/5
1/2 :1/5 = 2 va 1/2
3/8 = 1/2 -1/8
1/3:1/2 = 2/3
1/3:2 = 1/6
\(S=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
\(T=\dfrac{3}{1x2}+\dfrac{3}{2x3}+\dfrac{3}{3x4}+\dfrac{3}{4x5}+...\dfrac{3}{nx\left(n+1\right)}\)
\(T=3x\left[\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...\dfrac{1}{nx\left(n+1\right)}\right]\)
\(T=3x\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...\dfrac{1}{n}-\dfrac{1}{n+1}\right]\)
\(T=3x\left(1-\dfrac{1}{n+1}\right)=\dfrac{3xn}{n+1}\)
\(\frac{1}{1}+2+\frac{1}{1}+2+3+\frac{1}{1} +2+3+4+...+\frac{1}{1}+2+3......2019\)
Ta có : \(\frac{2}{2}+\left(\frac{1}{2}\right)+\frac{2}{2}+\left(1+2+3\right)+....+\frac{2}{2}+\left(1+2+....+50\right)\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{2550}\)
\(=\frac{2}{2}.3+\frac{2}{3}.4+....+\frac{2}{50}.51\)
\(=2.\left(\frac{1}{2}.3+\frac{1}{3}.4+.....+\frac{1}{50}.51\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2.\left(1-\frac{1}{51}\right)\)
\(=2.\frac{50}{51}\)
\(=\frac{100}{51}\)
Hmmm , kh bt có đúng kh nhỉ ???
Nếu kh đúng chỗ nào mong m.n chỉ ạ
:>>
cảm ơn mình đã k cho bạn rồi nhé cảm ơn