1-1/2+1/3-1/4+1/5-1/6+.....+1/99-1/100=1/51+1/52+1/53+1/54+..+1/100
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Chứng Minh:
1/1*2+1/3*4+1/5*6+...+1/97*98+1/99*100=1/51+1/52+1/53+...+1/99+1/100
Chứng tỏ: 1- 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +.........+ 1/99 - 1/100 = 1/51 + 1/52 + 1/53 + .....+ 1/100
1 - 1/2 + 1/3 - 1/4 +...+ 1/99 - 1/100
= (1 + 1/3 +...+ 1/99) - (1/2 + 1/4 +...+ 1/100)
= (1+1/2+1/3+...+1/100) - 2(1/2+1/4+...+1/100)
= (1+1/2+1/3+...+1/100) - (1+1/2+...+1/50)
= 1/51+1/52+...+1/100 (đpcm)
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Bạn đã được chuyển khoản số tiền 1.000.000.000 VND
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tính 1/51+1/52+1/53+....+1/100
1/1*2+1/3*4+1/5*6+...+1/99*100
Chứng minh :(1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+1/53+...+1/100
Chứng minh: 1- 1\2 + 1\3 - 1\4 + 1 \5 - 1\6 + ....... + 1\99 -1\100 = 1\51 + 1\52 + 1\53 + ..........+1\100
đây là j`? đầu đề hổng có, làm sao mà giải đc?????
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Xem thêm câu trả lời
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
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CMR(1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100):(1/51+1/52+1/53+...+1/100) = 1
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
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(1/51+1/52+1/53+...+1/100)÷(1/1×2+1/3×4+1/4×5+...+1/99×100)
tính c=1/1*2+1/3*4+1/5*6+...+1/99*100
tính b=1/51+1/52+1/53+...+1/100