\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)

\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)

\(S=2^2.\dfrac{2022}{2023}\)

\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)

x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023

x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023

x-(1-1/2023)=-2024/2023

x-2022/2023=-2024/2023

x = -2024/2023+2022/2023

x = -2/2023

Vậy x = -2/2023

=5(1-1/2+1/2-1/3+...+1/2023-1/2024)

=5*2023/2024

=10115/2024

\(=5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=5\left(1-\dfrac{1}{100}\right)\)

\(=5.\dfrac{99}{100}=\dfrac{99}{20}\)

Đầu tiên thì nhắc lại cái hằng đẳng thức cho bạn nào chưa học này: (a-b)²=a²-2ab+b²<=>a²+b²=(a-b)²+2ab

\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(=\dfrac{\left(\left(1-2\right)^2+2.1.2\right)}{1.2}+\dfrac{\left(\left(2-3\right)^2+2.2.3\right)}{2.3}+...+\dfrac{\left(\left(9-10\right)^2+2.9.10\right)}{9.10}\)
\(=\dfrac{\left(\left(-1\right)^2\right)}{1.2+2}+\dfrac{\left(\left(-1\right)^2\right)}{2.3+2}+...+\dfrac{\left(\left(-1^2\right)\right)}{9.10+2}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(=1-\dfrac{1}{10}+18\)
\(=18,9=\dfrac{189}{10}.\)

~ K chắc là đúng đâu ~

      $(\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{199.200}).x=\frac{2}{5}$

$\Rightarrow [9.(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200})]x=\frac{2}{5}$

$\Rightarrow [9.(\frac{1}{2}-\frac{1}{200})]x=\frac{2}{5}$

$\Rightarrow (9.\frac{99}{200})x=\frac{2}{5}$

$\Rightarrow \frac{891}{200}x=\frac{2}{5}$

$\Rightarrow$