\(a.x\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\Leftrightarrow x=-3\end{cases}}\)

\(b,\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(100x+\left(1+2+...+100\right)=5750\)

\(100x+\frac{100.101}{2}=5750\)

\(100x+5050=5750\)

\(100x=200\Leftrightarrow x=2\)

a) \(x.\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

b) (x + 1) + (x + 2) + ... + (x + 100) = 5750

(x + x + .... + x) + (1 + 2 + .. + 100) = 5750

                                   100x + 5050 = 5750

                                                 100x = 5750 - 5050

                                                  100x = 700

                                                        x = 700 : 100 = 7

B = (1 + 1/2)(1 + 1/3)(1 + 1/4) ...(1 + 1/100)

    = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{101}{100}\)

    = \(\frac{3.4.5....101}{2.3.4...100}=\frac{101}{2}\)

C = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{1000}\right)\)

   \(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)

    \(=\frac{1.2.3...999}{2.3.4....1000}=\frac{1}{1000}\)

ta có:\(n^2+1⋮n-1\)

\(\Rightarrow\left(n-1\right)^2+2=n^2+1\)

\(\Rightarrow n-1⋮n-1\)

vậy thỏa mãn yêu cầu bài ra

a/ -4 là B( a- 1 )

 Hay a - 1 là Ư(-4)

Ta có Ư(-4) = { -4; -2; -1; 1; 2 ; 4 }

Xét :

a - 1 = -4 => a = -3

a - 1 = -2 => a = -1

a - 1 = -1 => a = 0

a - 1 = 1 =>  a= 2

a - 1 = 2 => a = 3

a - 1 = 4 => a = 5

Ta có: \((2^3)^5.2^5\)

\(=2^{15}.2^5\)

\(=2^{15+5}=2^{20}\)

( 2³ )⁵ . 2⁵

= 2^3.5 . 2⁵

= 2¹⁵ . 2⁵

= 2^{15 + 5}

= 2²⁰

Vậy ...

a. -500

b.-35

c.-180

d. -1000

e. -27

g.-65

Phần e bn sai rồi,phải la -21