\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\div x=\frac{9}{10}\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\div x=\frac{9}{10}\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{20}\right)\div x=\frac{9}{10}\)

\(\Leftrightarrow\frac{19}{20}\div x=\frac{9}{10}\)

\(\Leftrightarrow x=\frac{19}{18}\)

Sửa đề : \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right):x=\frac{9}{10}\)

\(\Leftrightarrow VT=\frac{9}{10}x\)

\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)=\frac{9}{10}x\)

\(\Leftrightarrow\left(1-\frac{1}{20}\right)=\frac{9}{10}x\Leftrightarrow\frac{19}{20}=\frac{9}{10}x\)

\(\Leftrightarrow\frac{19}{20}=\frac{18x}{20}\) Khử mẫu ta đc : \(\Leftrightarrow18x=19\Leftrightarrow x=\frac{19}{18}\)

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

A =    1      +       1        +       1     +   ......   +     1                                                                                                                                             1x2          2x3              3x4                    19x20

A= (1/1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/19-1/20)

A= 1/1- 1/20

A= 19/20

Vậy A= 19/20

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

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

A = \(1-\frac{1}{20}\)

A = \(\frac{19}{20}\)

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

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

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

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

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

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

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

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

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