(1/a+1)+(1/a(a+1))=(a/a(a+1))+(1/a(a+1))=(a+1)/a(a+1)=1/a   

=>ĐPCM

l ike nhé

b)

\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(21.\frac{1}{2}\right)\)

\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}.\frac{76}{45}\right):\frac{21}{2}\)

\(\Rightarrow2-7< x< \left(1+\frac{38}{5}\right):\frac{21}{2}\)

\(\Rightarrow-5< x< \frac{43}{5}:\frac{21}{2}\)

\(\Rightarrow-5< x< \frac{86}{105}\)

Vì \(x\in Z\left(gt\right)\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}.\)

Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}.\)

Xét \(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}=\left(1+\frac{1}{2008}\right)+\left(\frac{1}{2}+\frac{1}{2007}\right)+...+\left(\frac{1}{1004}+\frac{1}{1005}\right)\)

\(=\frac{2009}{1.2008}+\frac{2009}{2.2007}+...+\frac{2009}{1004.1005}=2009.\left(\frac{1}{1.2008}+\frac{1}{2.2007}+...+\frac{1}{1004.1005}\right)\)

quy đồng mẫu số các phân số trong ngoặc: Gọi k₁ là thừa số phụ của \(\frac{1}{1.2008}\);...; k₁₀₀₄ là thừa số phụ của \(\frac{1}{1004.1005}\)

=> \(B=2009.\frac{k_1+k_2+...+k_{1004}}{1.2.3.4...2007.2008}\)

=> \(1.2.3....2007.2008.2009.\frac{k_1+k_2+...+k_{1004}}{1.2.3...2007.2008}=2009.\left(k_1+k_2+...+k_{1004}\right)\)

Tổng k₁ + k₂ + ...+ k₁₀₀₄ là số tự nhiên => A chia hết cho 2009 

de nay kho nhi

Bài 2 a:

\(A=n^3+3n^2+2n=n^3+n^2+2n^2+2n=n^2\left(n+1\right)+2n\left(n+1\right)=\left(n^2+2n\right)\left(n+1\right)=n\left(n+1\right)\left(n+2\right)\)

Mà tích 3 số nguyên liên tiếp chia hết cho 3,  suy ra A chia hết cho 3

\(\frac{A}{n}=\frac{4n+4}{n}=4+\frac{4}{n}\)
\(\Rightarrow n\in U\left(4\right)\)
Lập bảng tiếp nhé!
\(\frac{B}{n}=\frac{5n+6}{n}=5+\frac{6}{n}\)
Lập bảng

\(2.\)
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=\frac{3}{29}\cdot\frac{29}{3}-\frac{1}{5}\cdot\frac{29}{3}=1-\left(1+\frac{14}{15}\right)=1-1-\frac{14}{15}=\frac{14}{15}\)
b)\(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}=\frac{5}{9}\cdot\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)