C/m : 1 / 1.2 + 1/ 2.3 + 1/ 3.4 + ......... + 1 / 99.100 không là số nguyên
Tìm S=1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
Giải:
Ta có: \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{99.100}\)
\(\Leftrightarrow S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow S=\dfrac{1}{1}-\dfrac{1}{100}\)
\(\Leftrightarrow S=1-\dfrac{1}{100}\)
\(\Leftrightarrow S=\dfrac{99}{100}\)
Vậy ...
S= 1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
S=1-1/100=99/100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)
=1/1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100
=1-1/100
=99/100
=1−1/2+1/2−1/3+1/3−1/4+...+1/99−1/100
=1 − 1/100 = 99/100
1/1.2+1/2.3+1/3.4+...+1/99.100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
C=1 phần 1.2+1 phần 2.3+1 phần 3.4+...+1 phần 99.100
ta có
\(C=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{4.3}+..+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{99}{100}\)
\(=\frac{1}{100}\)
1/ Tập hợp A các số nguyên âm lớn hơn -100 là ?
2/Tính tổng S=1.2+2.3+3.4+.....+99.100 ta được kết quả S=?
1/ tap hop A co 99 phan tu {-1;-2;-3;......;-99}
2/ S=333300
Tính :
1/1.2 + 1/2.3 + 1/3.4 + . . . + 1/99.100
Answer:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{100}{100}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
1/1.2+1/2.3+1/3.4+1/4.5+...+1/99.100
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
1/(1.2)+1/(2.3)+1/(3.4)+...+1/(99.100)=?
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
1/(1.2)+1/(2.3)+1/(3.4)+...+1/(99.100)
=1-1/2+1/2-1-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
tôi không chép bài giang ho đai ca đâu nha.
=1-1/2+1/2-1/3+.....+1/99-1/100
=1-1/100
=99/100
tinh 1/1.2+1/2.3+1/3.4+.......+1/99.100
1/1.2+1/2.3+1/3.4+......+1/99.100
=1-1/2+1/2-1/3+1/3-1/4+..........+1/99-1/100
=1-1/100
=99/100
A=1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100
A=1/100-1
A=99/100