dạng tổng quát của mỗi phân số là 1/n(n+1) = 1/n -1/n+1

áp dụng vào làm với các phân số trong biểu thức cuối cùng còn 1-1/10=19/20

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)

\(A=1-\dfrac{1}{20}\)

\(A=\dfrac{19}{20}\)

A=\(\frac{19}{20}\),

sao ra vay ban minh muoc cach giai bai

Ta có A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{1}-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=1-\frac{1}{20}\)

\(A=\frac{19}{20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=1-\frac{1}{20}\)

\(A=\frac{19}{20}\)

Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
     => A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20

     =>A= 1-1/20=19/20

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)

\(\Rightarrow x=50\)

Vậy x = 50

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)

\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)

trả lời chỉ để lấy tích thời mọi người tích giùm hihi

bài này ở đâu z bạn

mình muốn xem bài này ở đâu để trả lời đừng hỏi tai sao

Đặt \(A=1.2+2.3+3.4+...+19.20\)

Ta có: \(A=1.2+2.3+3.4+...+19.20\)

        \(3A=1.2.3+2.3.3+3.4.3+...+19.20.3\)

        \(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+19.20.\left(21-1\right)\)

        \(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20\)

        \(3A=19.20.21\)

           \(A=19.20.7\)

           \(A=2660\)

\(1\cdot2+2\cdot3+...+19\cdot20=\frac{1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+...+19\cdot20\cdot\left(21-17\right)}{3}\)

   \(=\frac{1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+19\cdot20\cdot21-18\cdot19\cdot20}{3}\)\(=\frac{19\cdot20\cdot21}{3}=2660\)

của bạn Tạ Đức Hoàng Anh là đúng đấy