a, \(a^2\left(a+1\right)+2a\left(a+1\right)\)

\(=a\left(a+1\right)\left(a+2\right)\)

Vì \(a,a+1\) là 2 số tự nhiên liên tiếp nên:

\(\Rightarrow a\left(a+1\right)\) chia hết cho \(2\)

\(\Rightarrow a\left(a+1\right)\left(a+2\right)\) chia hết cho \(2\)

Vì \(a,a+1,a+2\) là 3 số tự nhiên liên tiếp nên:

\(\Rightarrow a\left(a+1\right)\left(a+2\right)\) chia hết cho 3

\(\Rightarrow a\left(a+1\right)\left(a+2\right)\) chia hết cho \(2.3\)

\(\Rightarrow a\left(a+1\right)\left(a+2\right)\) chia hết cho \(6\left(đpcm\right)\)

b, \(a\left(2a-3\right)-2a\left(a+1\right)\)

\(=a\left[2a-3-2\left(a+1\right)\right]\)

\(=-5a\) chia hết cho \(5\left(đpcm\right)\)

\(B=70\cdot\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)

\(B=70\cdot\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{10}\right)\right]\)

\(B=70\cdot13\cdot\frac{3}{70}\)

\(B=70\cdot\frac{3}{70}\cdot13\)

\(B=3\cdot13\)

\(B=39\)

a) (-1)^a =1 với a chẵn, (-1)^a =-1 với a lẻ

\(A=\left(-1\right)^{1+2+3+4+..+2010+2011}=\left(-1\right)^{\frac{2011+1}{2}.2011}=\left(-1\right)^{1006.2011}=1\)

Vì 1006 là số chẵn => 1006.2011 là số chẵn

b) \(B=70.\left(\frac{13.10101}{56.10101}+\frac{13.10101}{72.10101}+\frac{13.10101}{90.10101}\right)=70.\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)=3.13=39\)

c) Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

=> C=4

\(A=\left(-1\right)\cdot\left(-1\right)^2\cdot\left(-1\right)^3\cdot\left(-1\right)^4\cdot...\cdot\left(-1\right)^{2010}\cdot\left(-1\right)^{2011}\)

\(A=\left(-1\right)\cdot1\cdot\left(-1\right)\cdot1\cdot...\cdot1\cdot\left(-1\right)\)                \(\left(\text{Có 1006 số hạng }\left(-1\right)\right)\left(\text{Chẵn}\right)\) \(\Rightarrow\text{ }\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)=1\)

\(A=1\cdot1\cdot1\cdot...\cdot1\)

\(A=1\)

\(C=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{a+1}{a}\right):\dfrac{a-1}{a^3}\)

\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right):\dfrac{a-1}{a^3}\)

\(=\dfrac{a+2\cdot\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)

\(=\dfrac{2a\left(a+1\right)}{\left(a^2+1\right)\cdot\left(a+1\right)^3}\cdot\dfrac{a^2}{a-1}\)

\(=\dfrac{2a^3}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)

\(\frac{\left(1-2a\right)\left(1-2b\right)}{\left(1-a\right)\left(1-b\right)}-\frac{4\left(1-a-b\right)^2}{\left(2-a-b\right)^2}=\frac{\left(1-2a\right)\left(1-2b\right)\left(2-a-b\right)^2-4\left(1-a\right)\left(1-b\right)\left(1-a-b\right)^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)

\(=\frac{2a^3-2a^2b-3a^2-2ab^2+6ab+2b^3-3b^2}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)

\(=\frac{\left(2a^3-4a^2b+2ab^2\right)+\left(2a^2b-4ab^2+2b^3\right)-3\left(a^2-2ab+3b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)

\(=\frac{2a\left(a^2-2ab+b^2\right)+2b\left(a^2-2ab+b^2\right)-3\left(a^2-2ab+b^2\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)

\(=\frac{\left(a-b\right)^2\left(2a+2b-3\right)}{\left(1-a\right)\left(1-b\right)\left(2-a-b\right)^2}\)

\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)

\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)

\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)

\(=\dfrac{-4a^2-2a+1}{4a^2}\)