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 = \(-\frac{2015}{4028}\)

còn câu b thì sao z mấy bn?

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 \(abc\ne0\))

Lời giải:

\(A-6=5^1+5^2+...+5^{2015}\)

\(5(A-6)=5^2+5^3+...+5^{2016}\)

Trừ theo vế:
\(4(A-6)=5^{2016}-5^1\)

\(\Rightarrow A=\frac{5^{2016}-5}{4}+6=\frac{5^{2016}+19}{4}\)

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\(B=\frac{5^{1015}(5^{1001}+2)-10.5^{1014}-1}{4}=\frac{5^{2016}+2.5^{1015}-2.5^{1015}-1}{4}\)

\(=\frac{5^{2016}-1}{4}< \frac{5^{2016}+19}{4}\)

Do đó \(B< A\)

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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)

\(\Rightarrow M=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=1013+1=1014\)

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

\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)

\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)

\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)

\(D=\frac{1}{25}:13\)

\(D=\frac{1}{325}\)

`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`

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!