\(A=\frac{1}{2}.\left(\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)

\(\Rightarrow A=\frac{2}{4}\left(\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+.....+\frac{1}{51.55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)

\(VậyA=\frac{2}{55}\)

Câu 1: 

a: \(A=\dfrac{1}{2}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{55}=\dfrac{2}{55}\)

\(B=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-220}{18}=\dfrac{-110}{9}\)

\(A\cdot B=\dfrac{2}{55}\cdot\dfrac{-110}{9}=\dfrac{-4}{9}\)

Câu 2: 

a: |3-x|=x-5

=>|x-3|=x-5

\(\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left(x-5-x+3\right)\left(x-5+x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

\(A=\frac{2}{11\cdot15}+\frac{2}{15\cdot19}+...+\frac{2}{51\cdot55}\)

\(A=\frac{2}{4}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(A=\frac{1}{2}\cdot\frac{4}{55}\)

\(A=\frac{2}{55}\)

Bạn tham khảo bài làm của mình nhé !!

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}...+\frac{2}{51.55}\)

\(\Leftrightarrow2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{4}{55}\)

\(\Leftrightarrow A=\frac{4}{110}\)

ta phân tích như sau :

abcabc=abcx1001 vì 1001 chia hết cho 3 số nguyên 7 ;11;13 nên abcx1001cũng chia hết cho 7;11;13 mà abcabc=abcx1001 từ đó suy ra abcabc chia hết ít nhất 3 số nguyên tố

ta có:abcabc=abc.1001

mà 1001 chia hết cho 7;11;13(là số nguyên tố)

nên abc.1001 chia hết cho 7;11;13(là số nguyên tố)

suy ra số tự nhiên  abcabc  chia hết cho ít nhất 3 số nguyên tố

co: abcabc= abc . 1001

vì: 1001 chia hết cho 7; 11;13 (đều là các số nguyên tố)

=> abc . 1001 chia het cho 3 so nguyen to 7; 11; 13 

Vay moi so tu nhien co dang abcabc deu chia het cho it nhat 3 so nguyen to (DPCM)

\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)

\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

=1/2*4/55

=2/55