a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)

\(=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=\dfrac{100}{2}=50\)

b.\(B=\dfrac{x-2}{x+3}=\dfrac{x+3-5}{x+3}=1-\dfrac{5}{x+3}\)

\(\Rightarrow B\) có giá trị nguyên\(\Leftrightarrow\)\(1-\dfrac{5}{x+3}\) có giá trị nguyên.

\(\Rightarrow5⋮x+3\Leftrightarrow x+3\in\)\(U\left(5\right)=\left\{-5;-1;1;5\right\}\)

Ta có bảng:

Vậy x\(\in\left\{-8;-4;-2;2\right\}\)

la voi

đừng tick nó trong đó chả có j đâu

\(\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{100}{9.10}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.10}\)

\(=\dfrac{2.3.4.....10}{1.2.3......9}.\dfrac{2.3.4.....10}{3.4.5.....10}\)

\(=10.2\)

\(=20\)

\(2-4-6+8+10-12-14+16+...+1994-1996-1998+2000\)\(=\left(2-4-6+8\right)+\left(10-12-14+16\right)+...+\left(1994-1996-1998+2000\right)\)\(=0+0+...+0\)

\(=0\)