Question

[Nguyễn Đức Hùng]

Ai làm giúp mình   

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2022}+\dfrac{2}{x}=2\)

=>\(\dfrac{1}{2\text{x}\dfrac{3}{2}}+\dfrac{1}{3\text{x}\dfrac{4}{2}}+...+\dfrac{1}{2022\text{x}\dfrac{2023}{2}}+\dfrac{2}{x}=2\)

=>\(\dfrac{2}{2\text{x}3}+\dfrac{2}{3\text{x}4}+...+\dfrac{2}{2022\text{x}2023}+\dfrac{2}{x}=2\)

=>\(\dfrac{1}{2\text{x}3}+\dfrac{1}{3\text{x}4}+...+\dfrac{1}{2022\text{x}2023}+\dfrac{1}{x}=1\)

=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}+\dfrac{1}{x}=1\)

=>\(\dfrac{1}{2}-\dfrac{1}{2023}+\dfrac{1}{x}=1\)

=>\(\dfrac{1}{x}=1-\dfrac{1}{2}+\dfrac{1}{2023}=\dfrac{1}{2}+\dfrac{1}{2023}=\dfrac{2025}{4046}\)

=>\(x=\dfrac{4046}{2025}\)

Answer 2

[Nguyễn Thị Thương Hoài]

\(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+4+...+2022}\) + \(\dfrac{2}{x}\) = 2

\(\dfrac{1}{\left(1+2\right)\times2:2}\) + \(\dfrac{1}{\left(1+3\right)\times3:2}\)+ ... + \(\dfrac{1}{\left(1+2022\right)\times2022:2}\) + \(\dfrac{2}{x}\) = 2

\(\dfrac{2}{2\times3}\) + \(\dfrac{2}{3\times4}\) + ... + \(\dfrac{2}{2022\times2023}\) + \(\dfrac{2}{x}\) = 2

2 x (\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + ... + \(\dfrac{1}{2022\times2023}\)) + \(\dfrac{2}{x}\) = 2

2 \(\times\) (\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)) + \(\dfrac{2}{x}\) = 2

2 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{2023}\)) + \(\dfrac{2}{x}\) = 2

2 x \(\dfrac{1011}{2023}\) + \(\dfrac{2}{x}\)  = 2

    \(\dfrac{2021}{2023}\)  +  \(\dfrac{2}{x}\) = 2

                   \(\dfrac{2}{x}\) = 2 - \(\dfrac{2021}{2023}\)

                    \(\dfrac{2}{x}\) = \(\dfrac{2025}{2023}\)

                    \(x\) = 2 : \(\dfrac{2025}{2023}\)

                    \(x\) = \(\dfrac{4046}{2025}\)