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Toán Lớp 7: cho `A=1/(1.2)+1/(3.4)+1/(5.6)+…+1/(99.100)` `CMR : ` `1)A=1/51+1/52+…+1/99+1/100` `2)25/75+25/100

Home/ Toán Lớp 7: cho `A=1/(1.2)+1/(3.4)+1/(5.6)+…+1/(99.100)` `CMR : ` `1)A=1/51+1/52+…+1/99+1/100` `2)25/75+25/100

Toán Lớp 7: cho `A=1/(1.2)+1/(3.4)+1/(5.6)+…+1/(99.100)` `CMR : ` `1)A=1/51+1/52+…+1/99+1/100` `2)25/75+25/100
Phượng Tiên Tháng Sáu 10, 2022 Toán học 2 Comments 0 views

Toán Lớp 7: cho A=1/(1.2)+1/(3.4)+1/(5.6)+…+1/(99.100)
CMR :
1)A=1/51+1/52+…+1/99+1/100
2)25/75+25/100

Comments ( 2 )

doanthulan
10 Tháng Sáu, 2022 at 4:46 sáng

Reply

#chucbanhoctot
hienchaudiem
10 Tháng Sáu, 2022 at 4:46 sáng

Reply

a ) $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + … + $\frac{1}{99.100}$

= 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + … + $\frac{1}{99}$ – $\frac{1}{100}$

= ( 1 + $\frac{1}{2}$ + … + $\frac{1}{99}$ ) – ( $\frac{1}{2}$ + … + $\frac{1}{100}$ )

= ( 1 + $\frac{1}{2}$ + … + $\frac{1}{100}$ ) – 2 . ( $\frac{1}{2}$ + … + $\frac{1}{100}$ )

= ( 1 + $\frac{1}{2}$ + … + $\frac{1}{100}$ ) – ( 1 + … + $\frac{1}{50}$ )

= $\frac{1}{51}$ + $\frac{1}{52}$ + …. + $\frac{1}{100}$ ( đpcm )

b ) $\frac{25}{75}$ + $\frac{25}{100}$ = $\frac{1}{3}$ + $\frac{1}{4}$ = $\frac{7}{12}$ < A ( $\frac{99}{100}$ )

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