a: \(\left(2x-\dfrac{1}{5}\right)^2+\dfrac{16}{25}=\left(-2022\right)^0\)

=>\(\left(2x-\dfrac{1}{5}\right)^2+\dfrac{16}{25}=1\)

=>\(\left(2x-\dfrac{1}{5}\right)^2=1-\dfrac{16}{25}=\dfrac{9}{25}\)

=>\(\left[{}\begin{matrix}2x-\dfrac{1}{5}=\dfrac{3}{5}\\2x-\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{4}{5}\\2x=-\dfrac{2}{5}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

b: Sửa đề: \(\dfrac{x-1}{2024}+\dfrac{x-2}{2023}+\dfrac{x-3}{2022}+\dfrac{x-4}{2021}=4\)

=>\(\left(\dfrac{x-1}{2024}-1\right)+\left(\dfrac{x-2}{2023}-1\right)+\left(\dfrac{x-3}{2022}-1\right)+\left(\dfrac{x-4}{2021}-1\right)=4-4=0\)

=>\(\dfrac{x-2025}{2024}+\dfrac{x-2025}{2023}+\dfrac{x-2025}{2022}+\dfrac{x-2025}{2021}=0\)

=>x-2025=0

=>x=2025

Câu a:

\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} + \frac{16}{25} = \left(\right. - 2022 \left.\right)^{0}\)

Bước 1: Ta biết rằng:

\(\left(\right.-2022\left.\right)^0=1(\text{v}\overset{ˋ}{\imath}\text{ m}ọ\text{i s}\overset{ˊ}{\hat{\text{o}}}\text{ kh}\overset{ˊ}{\text{a}}\text{c 0 m}\overset{\sim}{\text{u}}\text{ 0 b}\overset{ˋ}{\overset{}{ă}}\text{ng 1})\)

Vậy ta có phương trình:

\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} + \frac{16}{25} = 1\)

Bước 2: Chuyển vế:

\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} = 1 - \frac{16}{25} = \frac{9}{25}\)

Bước 3: Lấy căn hai vế:

\(2 x - \frac{1}{5} = \pm \frac{3}{5}\)

TH1:

\(2 x - \frac{1}{5} = \frac{3}{5} \Rightarrow 2 x = \frac{4}{5} \Rightarrow x = \frac{2}{5}\)

TH2:

\(2 x - \frac{1}{5} = - \frac{3}{5} \Rightarrow 2 x = - \frac{2}{5} \Rightarrow x = - \frac{1}{5}\)

✅ Đáp số câu a: \(x = \frac{2}{5}\) hoặc \(x = - \frac{1}{5}\)

Câu b:

\(\frac{x - 1}{2024} + \frac{x - 2}{2023} + \frac{x - 3}{2022} + \frac{x - 4}{2021}\)

Đây không phải phương trình (không có dấu bằng), nên mình hiểu bạn đang yêu cầu rút gọn biểu thức này.

Tách từng tử riêng ra:

\(= \frac{x}{2024} - \frac{1}{2024} + \frac{x}{2023} - \frac{2}{2023} + \frac{x}{2022} - \frac{3}{2022} + \frac{x}{2021} - \frac{4}{2021}\)

Gom nhóm các phần x lại:

\(= x \left(\right. \frac{1}{2024} + \frac{1}{2023} + \frac{1}{2022} + \frac{1}{2021} \left.\right) - \left(\right. \frac{1}{2024} + \frac{2}{2023} + \frac{3}{2022} + \frac{4}{2021} \left.\right)\)

✅ Vậy biểu thức rút gọn là:

\(x \left(\right. \frac{1}{2024} + \frac{1}{2023} + \frac{1}{2022} + \frac{1}{2021} \left.\right) - \left(\right. \frac{1}{2024} + \frac{2}{2023} + \frac{3}{2022} + \frac{4}{2021} \left.\right)\)

a, (2x -1/5)^2 + 16/25 = ( -2022) ^0

(2x - 1/5 ) ^2 +16/25 = 1

(2x - 1/5 ) ^2 = 1 -16/25

(2x - 1/5 ) ^2 = 9/25

(2x - 1/5 ) ^2 = (3/5)^2

2x - 1/5 = 3/5

2x = 3/5 + 1/5

2x = 4/5

x = 4/5 : 2

x = 2/5

b, Ta có: \(\frac{x + 4}{2021} + \frac{x + 3}{2022} = \frac{x + 2}{2023} + \frac{x + 1}{2024}\)

=>\(\left(\right. \frac{x + 4}{2021} + 1 \left.\right) + \left(\right. \frac{x + 3}{2022} + 1 \left.\right) - \left(\right. \frac{x + 2}{2023} + 1 \left.\right) - \left(\right. \frac{x + 1}{2024} + 1 \left.\right) = 0\)

=>\(\frac{x + 2025}{2021} + \frac{x + 2025}{2022} - \frac{x + 2025}{2023} - \frac{x + 2025}{2024} = 0\)

=>x+2025=0

=>x=-2025