\(\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
Những câu hỏi liên quan
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tính:
\(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)
Tính nhanh :
A = \(\left(\frac{2}{3}+\frac{3}{4}+....+\frac{99}{100}\right)\cdot\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....+\frac{98}{99}\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\cdot\left(\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}\right)\)
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
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26 tháng 3 2016 lúc 16:34
Tính dãy số sau :
\(D=\frac{100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}+\frac{99}{100}}\)
Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
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)
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\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
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Afrac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-...+frac{99}{3^{99}}-frac{100}{3^{100}}Rightarrow3A1-frac{2}{3}+frac{3}{3^2}-...+frac{99}{3^{98}}-frac{100}{3^{99}}Rightarrow3A+Aleft(...right)+left(...right)Rightarrow4A1-frac{1}{3}+frac{1}{3^2}-...-frac{1}{3^{99}}-frac{100}{3^{100}}Rightarrow3.4A3-1+frac{1}{3}-...-frac{1}{3^{98}}-frac{100}{3^{99}}Rightarrow12A+4Aleft(...right)+left(...right)Rightarrow16A3-frac{101}{3^{99}}-frac{100}{3^{100}} 3Rightarrow A frac{3}{16}
Đọc tiếp
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(...\right)+\left(...\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow12A+4A=\left(...\right)+\left(...\right)\)
\(\Rightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}< 3\)
\(\Rightarrow A< \frac{3}{16}\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
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tính nhanh
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
CMR:
a,\(100\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+........+\frac{99}{100}\)
\(VP=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(VP=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
\(VP=\frac{2}{2}-\frac{1}{2}+\frac{3}{3}-\frac{1}{3}+\frac{4}{4}-\frac{1}{4}+...+\frac{100}{100}-\frac{1}{100}\)
\(VP=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)
\(VP=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=VT\) ( đpcm )
Mk nghĩ \(VT=100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\) bn xem lại đề có nhầm ko
Chúc bạn học tốt ~
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tính nhanh: \(\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
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
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