\(\left(15.3-21\right):4+108-145\)

\(=\left(45-21\right):4+108-145\)

\(=24:4+108-145\)

\(=6+108-145\)

\(=114-145=-31\)

(15.3-21):4+108-145 

=  24:4 + 108-145

= 6 + 108-145

= -31

\(\dfrac{15}{4}-\dfrac{11}{7}\times\dfrac{14}{33}=\dfrac{15}{4}-\dfrac{11\times14}{7\times33}=\dfrac{15}{4}-\dfrac{2}{3}=\dfrac{45}{12}-\dfrac{8}{12}=\dfrac{37}{12}\)

15\4 - 11\7 x 14\33

=15/4 - 11x14 / 7x33

= 15/4 - 154/231

= 15/4 - 154: 77/231:77

= 15/4 - 2/3

= 15 x 3 / 4x 3 - 2x4 / 3x4

= 45/12 - 8/12

= 37/12

a) \(\dfrac{3}{4}x=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}:\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

b) \(x-\dfrac{4}{7}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}+\dfrac{4}{7}\)

\(\Leftrightarrow x=\dfrac{29}{14}\)

\(B=1^2+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+100\left(99+1\right)=\)

\(=\left(1+2+3+...+100\right)+\left(1.2+2.3+3.4+...+99.100\right)=\)

1+2+3+...+100 tổng cấp số cộng d=1

\(1.2+2.3+3.4+...+99.100=\)

\(=\dfrac{1}{3}\left(1.2.3+2.3.3+3.4.3+...+99.100.3\right)=\)

\(=\dfrac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)}{3}=\)

\(=\dfrac{1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-.....-98.99.100+99.100.101}{3}=\)

\(=\dfrac{99.100.101}{3}=33.100.101\)

Thay các kết quả vào để tính B

\(x=0-\dfrac{1}{2}\)

\(x=-\dfrac{1}{2}\)

x+1/2 = 0

= x = 0-1/2

= x = -1/2

Ta có: \(n+2⋮n-1\)

\(\Leftrightarrow n-1+3⋮n-1\)

\(\Leftrightarrow3⋮n-1\)

Vì \(n\in Z\)\(\Rightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow n\in\left\{2;0;4;-2\right\}\)

Vậy....

\(n+2⋮n-1\)

<=> (n-1) +3 \(⋮\) n-1

Vì n-1\(⋮\) n-1 

Để n+2\(⋮\) n-1 => 3\(⋮\) n-1 hay n-1\(\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

=> n \(\in\left\{2;4;0;-2\right\}\)

Giúp mình vs ạ