\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
K
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HT
2 tháng 9 2015
A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
A = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)...\left(-\frac{2013.2015}{2014.2014}\right)\)
A = \(-\left[\frac{\left(1.2....2013\right)\left(3.4....2015\right)}{\left(2.3....2014\right)\left(2.3...2014\right)}\right]\)
A = \(-\left(\frac{2015}{2014.2}\right)\)
A = \(-\frac{2015}{4028}\)
BA
17 tháng 11 2019
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
AH
Akai Haruma
Giáo viên
22 tháng 8 2019
Lời giải:
\(A-6=5^1+5^2+...+5^{2015}\)
\(5(A-6)=5^2+5^3+...+5^{2016}\)
Trừ theo vế:
\(4(A-6)=5^{2016}-5^1\)
\(\Rightarrow A=\frac{5^{2016}-5}{4}+6=\frac{5^{2016}+19}{4}\)
--------------
\(B=\frac{5^{1015}(5^{1001}+2)-10.5^{1014}-1}{4}=\frac{5^{2016}+2.5^{1015}-2.5^{1015}-1}{4}\)
\(=\frac{5^{2016}-1}{4}< \frac{5^{2016}+19}{4}\)
Do đó \(B< A\)
HC
1
24 tháng 11 2021
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NT
1
14 tháng 8 2017
Đặt a=2013
\(\Rightarrow M=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{a^2+2a+1+a^4+2a^3+a^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{\left(a^4+2a^3+a^2\right)+2\left(a^2+a\right)+1}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\left(\frac{a^2+a+1}{a+1}\right)^2}+\frac{a}{a+1}\)
\(\Rightarrow M=\frac{a^2+a+1+a}{a+1}\)(Bỏ trị tuyệt đối vì a=2013)
\(\Rightarrow M=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=1013+1=1014\)
S
28 tháng 4 2016
D= [(1-1/2)(1-1/3)...(1-1/25)]:[(1+1/2)(1+1/3)...(1+1/25)]
D= [1/2. 2/3. ... . 24/25]: [3/2. 4/3. ... . 26/25]
D= 1/25 : 2/26
D= 1/25 . 26/2= 13/25
Vậy D= 13/25
28 tháng 4 2016
\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)
\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)
\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)
\(D=\frac{1}{25}:13\)
\(D=\frac{1}{325}\)
13 tháng 6 2021
`A=(10^14-1)/(10^15-11)`
`=>10A=(10^15-10)/(10^15-11)`
`=>10A=(10^15-11+1)/(10^15-11)`
`=>10A=1+1/(10^15-1)`
`=>A>1/10`
`B=(10^14+1)/(10^15+9)`
`=>10B=(10^15+10)/(10^15+9)`
`=>10A=(10^15+9+1)/(10^15+9)`
`=>10A=1+1/(10^15+9)`
Vì `1/(10^15-1)>1/(10^15+9)`
`=>10B>10A`
`=>B>A`
13 tháng 6 2021
Giải:
\(A=\dfrac{10^{14}-1}{10^{15}-11}\)
\(10A=\dfrac{10^{15}-10}{10^{15}-11}\)
\(10A=\dfrac{10^{15}-11+1}{10^{15}-11}\)
\(10A=1+\dfrac{1}{10^{15}-11}\)
Tương tự:
\(B=\dfrac{10^{14}+1}{10^{15}+9}\)
\(10B=\dfrac{10^{15}+10}{10^{15}+9}\)
\(10B=\dfrac{10^{15}+9+1}{10^{15}+9}\)
\(10B=1+\dfrac{1}{10^{15}+9}\)
Vì \(\dfrac{1}{10^{15}-11}>\dfrac{1}{10^{15}+9}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Chúc bạn học tốt!
\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)
\(=0\)