Bài 2:

\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)

Bài 1:

\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)

Lời giải:
Gọi giá của tivi ban đầu là $a$ đồng. 

Giá tivi đầu tháng: $a(1+0,05)=1,05a$ (đồng) 

Giá tivi của tháng: $1,05a(1-0,05)=0,9975a$ (đồng) 

Chênh lệch giá ti vi ban đầu và cuối tháng là:
$a-0,9975a=50000$

$a.0,0025=50000$

$a=20000000$ (đồng)

41........ in listening to pop music

42 ........ small for my daughter to wear

43 ........... largest country in the world

44 ........ the owner of that house

45 ........ necessary to do exercise regularly

Bài 3:

Để A là số nguyên thì \(2n+7⋮n+3\)

=>\(2n+6+1⋮n+3\)

=>\(1⋮n+3\)

=>\(n+3\in\left\{1;-1\right\}\)

=>\(n\in\left\{-2;-4\right\}\)

Bài 4:

Để A là số nguyên thì \(2n+7⋮n+1\)

=>\(2n+2+5⋮n+1\)

=>\(5⋮n+1\)

=>\(n+1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{0;-2;4;-6\right\}\)

Bài 5:

Để A là số nguyên thì \(6n-3⋮3n+1\)

=>\(6n+2-5⋮3n+1\)

=>\(-5⋮3n+1\)

=>\(3n+1\in\left\{1;-1;5;-5\right\}\)

=>\(3n\in\left\{0;-2;4;-6\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

mà n nguyên 

nên \(n\in\left\{0;-2\right\}\)

Bài 6:
Để A là số nguyên thì \(3n+4⋮n-1\)

=>\(3n-3+7⋮n-1\)

=>\(7⋮n-1\)

=>\(n-1\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{2;0;8;-6\right\}\)