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A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)
ta cho nó dài hơn như sau
A=(2/3+3/4+4/5+5/6+....+98/99+99/100)
ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số
2:3:4:5...99 vậy ta còn các số 2/100
ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99
làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100
vậy với (2/3+3/4+...+98/99) ra 2/99
xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây
\(\frac{0,8:\left(\frac{4}{5}:1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(100-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{27}}+\left(1,2.0,5\right):\frac{3}{5}\)
=\(\frac{0,8:0,64}{0,64.25-\frac{1}{25}}+\frac{\left(100-\frac{2}{5}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{56}{27}}+\left(1,2.0,5\right).\frac{5}{3}\)
=
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
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)
\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)