CMR (1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+....+ 1/100)=1/ 51+1/52+....+1/99+1/100
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
CMR: 1/51 + 1/52 + 1/52 +...+1/100 = 1-1/2 + 1/3 - 1/4 +...+1/99-1/100
Ta có \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Rightarrow\text{Đ}PCM\)
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
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
Xét VT:
\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(VT=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=VP\)
=>đpcm
Ta xét vế trái:
\(vt=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(VT=VP\)
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)
Bạn đã được chuyển khoản số tiền 1.000.000.000 VND
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?????
chứng minh rằng:
1-1/2+1/3-1/4+1/5-1/6+....+1/99-1/100=1/51+1/52+....+1/100