So sánh A và B :
A=1/31 + 1/32 + 1/33+....+1/60
B=1/1.2+1/3.4+1/5.6+.....1/59.60
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14 tháng 8 2016
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\)
\(B=\left(1+\frac{1}{3}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{59}+\frac{1}{60}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)\)
\(B=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=A\)
11 tháng 8 2016
B = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/59.60
B = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/59 - 1/60
B = (1 + 1/3 + 1/5 + ... + 1/59) - (1/2 + 1/4 + 1/6 + ... + 1/60)
B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - 2.(1/2 + 1/4 + 1/6 + ... + 1/60)
B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - (1 + 1/2 + 1/3 + ... + 1/30)
B = 1/31 + 1/32 + 1/33 + ... + 1/60 = A
=> B = A
DD
11 tháng 8 2016
ta có: Lớn nhất của A là:\(\frac{1}{31}+\frac{1}{31}+...+\frac{1}{31}\)(30 phân số)
=30/31
B=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{3}+...+\frac{1}{59}-\frac{1}{60}\)\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)
Bé nhất của của B là :\(\left(1+1+...+1\right)-\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(=30-\frac{30}{60}\)
=>B>A
NT
1
ML
14 tháng 8 2015
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)
=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}=\left(1+\frac{1}{3}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{59}+\frac{1}{60}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{60}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)
11 tháng 4 2017
Đặt: \(\left\{{}\begin{matrix}A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\end{matrix}\right.\)
Ta có:
\(B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{59}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\)
\(\Rightarrow B=A\)
Vậy \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\) (Đpcm)
TM
11 tháng 4 2017
Ta có:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+......+\dfrac{1}{59.60}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{59}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)
- \(2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{60}\right)\) - \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)+ \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)
- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)
= \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)
Vậy\(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{59.60}\)
VM
1
YN
26 tháng 5 2021
\(P=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)
\(\text{Ta có:}\)
\(91=13.7\)
\(\rightarrow4.13+5.17=42.35⋮91\)
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)
\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.....60.42.35\)
\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32....60.20.91⋮91\)
O
1
S
22 tháng 4 2018
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{59\cdot60}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{69}-\frac{1}{60}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
KT
27 tháng 3 2018
\(A=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right):2\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right):2\)
\(=\left(1-\frac{1}{2017}\right):2\)\(< \)\(\frac{1}{2}\) (Do 1 - 1/2017 < 1)
\(\text{Có 3 trường hợp có thể xảy ra:}\)
\(A=B\)
\(A< B\)
\(A>B\)
mik cần giải mà