Đặt \(x=\dfrac{1}{49\cdot44}+\dfrac{1}{44\cdot39}+...+\dfrac{1}{14\cdot9}+\dfrac{1}{9\cdot4}\) và y = ... (thừa số thứ hai chưa ghi rõ, nếu ghi rõ thì mới làm được)

Ta có:

\(5x=5\left(\dfrac{1}{49\cdot44}+\dfrac{1}{44\cdot39}+...+\dfrac{1}{14\cdot9}+\dfrac{1}{9\cdot4}\right)\)

\(5x=\dfrac{5}{49\cdot44}+\dfrac{5}{44\cdot39}+...+\dfrac{5}{14\cdot9}+\dfrac{5}{9\cdot4}\)

\(5x=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{39}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{49}\)

\(5x=\dfrac{1}{4}-\dfrac{1}{49}=\dfrac{45}{196}\)

\(x=\dfrac{45}{196}\div5=\dfrac{9}{196}\)

Từ đây tự tìm y (thừa số thứ hai)

Suy ra \(A=xy=\dfrac{9}{196}\cdot...=...\)

Cho em hỏi là A - B thì phải là ( 2²⁰¹⁸ - 1 ) - 2²⁰¹⁸ chứ đúng không ạ?

       A =  1 + 2 +  2²+...+ 2²⁰¹⁷

     2A =       2  + 2² +...+2²⁰¹⁷ + 2²⁰¹⁸

2A - A =  2²⁰¹⁸ -  1

        A = 2²⁰¹⁸ - 1

P = A - B =  ( 2²⁰¹⁸ - 1) = 2²⁰¹⁸ - 1 - 2²⁰¹⁸  =  - 1

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}-\dfrac{3}{2}+1=\)

\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\)

= 4 . -1/8 - 2 . -1/4 + 3 . -1/2 + 1

= -1/2 - -1/2 + -3/2 + 1

= -1/2

Xét tam giác ABC, có:ˆA1+ˆB+ˆC1=1800 (Tổng các góc của tam giác)Xét tam giác ADC, có:ˆA2+ˆD+ˆC2=1800 (Tổng các góc của tam giác)Xét tứ giác ABCD, có:ˆA+ˆB+ˆD+ˆC=ˆA1+ˆA2+ˆB+ˆD+ˆC1+ˆC2=(ˆA1+ˆB+ˆC1)+(ˆA2+ˆD+ˆC2)=1800+1800=3600⇒đpcm

\(3^8.9^2=3^8.\left(3^2\right)^2=3^8.3^4=3^{8+4}=3^{12}\)

Lần sau nên chọn đúng môn.

\(\dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\left(\dfrac{0,15}{0,5}\right)^4.\dfrac{1}{0,5}=\left(\dfrac{3}{10}\right)^4.2=\dfrac{81}{10000}.2=\dfrac{81}{5000}\)

KQ = 81/5000

\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)

        \(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) 

=>      \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\) 

=>        \(2x+\dfrac{1}{2}=\pm\dfrac{1}{2}\) 

TH1: 

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

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

          \(x=0\) 

TH2:

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

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

        \(2x=-1\) 

            \(x=\dfrac{-1}{2}\) 

      Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)

\(\dfrac{5}{7}:x+1=\dfrac{-15}{30}\)

\(\dfrac{5}{7}:x=\dfrac{-15}{30}-1\)

\(\dfrac{5}{7}:x=\dfrac{-45}{30}\)

\(x=\dfrac{5}{7}:\dfrac{-45}{30}\)

\(x=\dfrac{-10}{21}\)

\(\dfrac{5}{7}:x+1=-\dfrac{15}{30}\)

\(\Rightarrow\dfrac{5}{7}:x+1=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{1}{2}-1\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{5}{7}:-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{10}{21}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{9}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=3\)

\(\Rightarrow x=3+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{7}{2}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\\ \\ \\ \Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}=\dfrac{9}{3}=3\\ \\ \\ \Rightarrow x=3+\dfrac{1}{2}=3\dfrac{1}{2}\)