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b) \(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100.99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)
\(=\dfrac{1}{100.99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{100.99}-\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{100.99}-\dfrac{98}{99}=-\dfrac{9799}{9900}\)
1/2+2/5+1/3+5/7+1/6+(-4/35)+1/41
đổi ra rồi nha bạn thế là tự biết làm được rồi
khó nhìn lắm bn ak
sao pn ko cho
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{8}+\frac{4}{9}+\frac{17}{14}.\)
thì có phải dễ nhìn hơn ko
a, \(A=\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\frac{11}{125}+\left(\frac{-17}{18}+\frac{4}{9}\right)+\left(\frac{-5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}+\frac{-1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)
b, \(B=1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
\(=\left(1+2+3+4-3-2-1\right)-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=4-3=1\)
c, \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-49}{50}\)
Nguyễn Đăng Duy ơi bài trên là tính nhanh hay tính vậy bạn .
c/
C = 1/100-1/100-1/99-1/99-1/98-1/98-1/97-..........-1/3-1/2-1/2-1/1
C = 1/100-1/100-1/1
C = 0-1/1
C = -1
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1
C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)
C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)
C = 1/100 - (1 - 1/100)
C = 1/100 - 99/100
C = -98/100 = -49/50
\(\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\frac{99}{100}\)
\(=\frac{-49}{50}\)
