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Nhân S với 2 ta được:
S = 2/1x2x3 + 2/2x3x4 + 2/3x4x5 + ... + 2/98x99x100
= (1/1x2 – 1/2x3) + (1/2x3 – 1/3x4) + (1/3x4 – 1/4x5) + …….. + (1/98x99 – 1/99x100)
= 1/1x2 – 1/99x100 = 1/2 – 1/9900 = 9898/19800
Vậy:
S = 1/1x2x3 + 1/2x3x4 + 1/3x4x5 + ... + 1/98x99x100
= 9898/19800 : 2
S = 4949/19800
Bài này khi sáng mình mới học 100% là đúng luôn.
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ........... + 1/99x100.
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..........1/98-1/99+1/99-1/100.
=1/1-1/100=99/100.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=+....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
= \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=1\)
ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
bài toán được viết lại như sau:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\)
vậy x=2012
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)\(\frac{1}{100}\)
A = \(1-\frac{1}{100}\)
A = \(\frac{100}{100}-\frac{1}{100}\)
A = \(\frac{99}{100}\)
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
a) \(\left(x-25\right):15=20\)
\(\Rightarrow x-25=20\times15\)
\(\Rightarrow x-25=300\)
\(\Rightarrow x=300+25\)
\(\Rightarrow x=325\)
Vậy x = 325
b) \(3\times x-25=80\)
\(\Rightarrow3\times x=80+25\)
\(\Rightarrow3\times x=105\)
\(\Rightarrow x=105:3\)
\(\Rightarrow x=35\)
Vậy x = 35
c) \(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{2}-\frac{1}{100}\)
\(S=\frac{49}{100}\)
Vậy \(S=\frac{49}{100}\)
_Chúc bạn học tốt_
Mình chỉ tính câu b và c thội nhé!.
Ta có:
b) \(1.2+2.3+3.4+...+99.100\)
\(=\frac{99.100.101}{3}=333300\)
c) \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{1+1+2+1+2+3+1+2+3+4+...+1+2+3+...+99}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+....+99}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)
mình nghĩ bạn điền sai đề
ukm mk đánh sai thiệt phải là 1/5x6 chứ ko phải 1/4x5