Tính tổng một cách hợp lí :
E= \(\frac{1}{100\cdot99}\) - \(\frac{1}{99\cdot98}\) - \(\frac{1}{98\cdot97}\) - .......-\(\frac{1}{3\cdot2}\) - \(\frac{1}{2\cdot1}\)
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Tính nhanh:
C=\(\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
bài này dễ lắm,mình giải đây:
C = \(\frac{1}{100}\)- \(\frac{1}{100.99}\)-\(\frac{1}{99.98}\)\(\frac{1}{98.97}\)- ... - \(\frac{1}{3.2}\)- \(\frac{1}{2.1}\)
C = \(\frac{-1}{1.2}\)+ \(\frac{-1}{2.3}\) + ... +\(\frac{-1}{98.99}\)+ \(\frac{1}{99.100}\)+ \(\frac{1}{100}\)
C = \(\frac{-1}{1}\)- \(\frac{-1}{2}\)
Mình bận rồi , phần sau tự làm nha.
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tính
\(\frac{1}{100\cdot99}\)-\(\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Tính :
\(A=\frac{1\cdot98+2\cdot97+3\cdot96+......+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+......+98\cdot99}\)
\(B=\frac{100-\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.........+\frac{99}{100}}\)
\(\frac{1\cdot98+2\cdot97+3\cdot96+...+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+98\cdot99}\)
Giúp mình nhé mình đang cần gấp
\(\frac{1\cdot98+2\cdot97+3\cdot96+...+96\cdot3+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+96\cdot97+97\cdot98+98\cdot99}\)
thực hiện phép tính
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{2}-1\right)\)
\(=\frac{1}{100}-\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}+1\)
\(=\frac{9799}{9900}\)
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\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Gọi A=\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
A= -(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\))
A=-(1-\(\frac{1}{100}\))
A=-(\(\frac{99}{100}\))
A=-99/100
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\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Leftrightarrow-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(\Leftrightarrow\)\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow-\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow-\left(\frac{99}{100}\right)\)
\(=-\frac{99}{100}\)
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Đặt \(A=\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Leftrightarrow A=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(\Leftrightarrow A=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow A=-\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow A=-\frac{99}{100}\)
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tính:\(\frac{1\cdot98+2\cdot97+3\cdot96+...+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}\)
tính
\(\frac{1\cdot98+2\cdot97+3\cdot96+...+96\cdot3+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+96\cdot97+97\cdot98+98\cdot99}\)
ai đó giúp mk mha mk sẽ tick cho người giúp mk làm ra đầu tiên