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XO
18 tháng 10 2020
Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
25 tháng 8 2023
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
HT
1
MA
12 tháng 11 2016
ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{b+a}{ab}\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
tíc mình nha
CE
0
DT
0
NQ
4
NB
1
NG
28 tháng 3 2020
Đáp án:
Cho a,b,c thỏa mãn:
2ab(2b-a)-2ac(c-2a)-2bc(b-2c)= 7abc
CMR:Tồn tại 1số bằng 2 số kia.
Giải thích các bước giải:
VC
0
26 tháng 8 2019
a, \(A=\frac{1}{2}+\left[\frac{1}{2}\right]^2+\left[\frac{1}{2}\right]^3+...+\left[\frac{1}{2}\right]^{99}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right]-\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right]\)
\(A=1-\frac{1}{2^{99}}\)
Do đó A < 1
b, \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(3B-B=\left[1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]-\left[1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right]\)
\(2B=1-\frac{1}{3^{99}}\)
\(B=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
6 tháng 9 2021
a) \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1=a^3+a^2b-a^2b+b^3+1=a^3+b^3+1\)
b) \(P=x\left(x-y+1\right)-y\left(y+1-x\right)-2=x^2-xy+x-y^2-y+xy-2=x^2+x-y-y^2-2\)
c) \(Q=\left(m+3\right)\left(m^2+3m-5\right)+\left(6-m\right)m^2+11=m^3+3m^2-5m+3m^2+9m-15+6m^2-m^3+11=12m^2+4m-4\)
6 tháng 9 2021
a: Ta có: \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1\)
\(=a^3+a^2b-a^2b+b^3+1\)
\(=a^3+b^3+1\)
a, đề phải là 1/a.(a+1) = 1/a - 1/a+1 chứ bạn !
Có : 1/a.(a+1) = (a+1)-a/a.(a+1) = a+1/a.(a+1) - a/a.(a+1) = 1/a - 1/a+1
=> 1/a.(a+1) = 1/a - 1/a+1
b, Có : 2/a.(a+1).(a+2) = (a+2)-a/a.(a+1).(a+2) = a+2/a.(a+1).(a+2) - a/a.(a+1).(a+2) = 1/a.(a+1) - 1/(a+1).(a+2)
=> 2/a.(a+1).(a+2) = 1/a.(a+1) - 1/(a+1).(a+2)
Tk mk nha
a, \(VP=\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a\left(a+1\right)}==\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}=VT\)
b, \(VP=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}=VT\)