2C

3C

4A

5C

6D

7B

Sai. Vì: \(98dm^2=0,98m^2\) 

Khối lượng m có giá trị: \(m=7,8.V=7,8.2=15,6\left(g\right)\) 

1350; 3873; 17856; 4911

Số chia hết cho 9: 4599; 10539; 18990; 36270

\(\displaystyle \begin{array}{{>{\displaystyle}l}} xy-5x+y=11\ \Leftrightarrow \ y( x+1) -5( x+1) =6\\ \Leftrightarrow \ ( x+1)( y-5) =6\\ Do\ x,y\ \in Z\ nên\ ta\ có:\\ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x+1 & \ \ \ 1\ \ \ & \ \ -1\ \ & \ \ \ 6\ \ \ & \ \ -6\ \ & \ \ \ 2\ \ \ & \ \ -2\ \ & \ \ \ 3\ \ \ & \ \ -3\ \ \\ \hline y-5 & \ 6 & -6 & \ 1 & \ \ -1 & 3 & -3 & 2 & -2\\ \hline x & 0 & -2 & 5 & -7 & 1 & -3 & 2 & -4\\ \hline y & 11 & -1 & 6 & 4 & 8 & 2 & 7 & 3\\ \hline \end{array} \ \ \ \\ Thử\ lại\ thấy\ các\ cặp\ giá\ trị\ ( x;y) \ trên\ đều\ thỏa\ mãn\\ Vậy\ ( x;y) \in \{( 0;11) ;( -2;-1) ;( 5;6) ;( -7;4) ;( 1;8) ;( -3;2) ;( 2;7) ;( -4;3)\} \ \end{array}\)

\( \begin{array}{l} \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{6} +\frac{1}{12} +\frac{1}{20} +...+\frac{1}{240}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2.3} +\frac{1}{3.4} +\frac{1}{4.5} +...+\frac{1}{15.16}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{3} +\frac{1}{3} -\frac{1}{4} +...+\frac{1}{15} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\frac{7}{16}\\ \Leftrightarrow \ B=\frac{7}{32} \end{array}\)

\( \begin{array}{l} \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{6} +\frac{1}{12} +\frac{1}{20} +...+\frac{1}{240}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2.3} +\frac{1}{3.4} +\frac{1}{4.5} +...+\frac{1}{15.16}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{3} +\frac{1}{3} -\frac{1}{4} +...+\frac{1}{15} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\frac{7}{16}\\ \Leftrightarrow \ B=\frac{7}{32} \end{array}\)

Ta có:

 \( \begin{array}{l} a+b=5\ \ \ \ \ \ ( 1)\\ b+c=16\ \ \ \ ( 2)\\ c+a=19\ \ \ \ ( 3)\\ Từ\ ( 1) \ và\ ( 2) ,\ trừ\ vế\ theo\ vế,\ ta\ được:\ a-c=-11\\ \Leftrightarrow \ c-a=11\ \ \ ( 4)\\ Từ\ ( 3) \ và\ ( 4) ,\ cộng\ vế\ theo\ vế,\ ta\ được:\ 2c=30\ \Leftrightarrow \ c=15\\ Khi\ đó:\ a+b-c=5-15=-10\ \end{array}\)

Từ 15 điểm phân biệt, ta tạo được: \(\dfrac{15.\left(15-1\right)}{2}=105\) (đoạn thẳng)