\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)

\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)

a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

a)\(5x^5=x^3\Leftrightarrow5x^5-x^3=0\Leftrightarrow x^3\left(5x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{\sqrt{5}}\\x=-\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)

Vì \(x\in N\Rightarrow x=0\)

b)\(\left(2x-3\right)^3=64\Leftrightarrow\left(2x-3\right)^3=4^3\)

\(\Rightarrow2x-3=4\Rightarrow x=\dfrac{7}{2}\) loại vì \(x\in N\)

c)\(\left(5x-1\right)^2=2023^0\cdot7+3^2\)

\(\Rightarrow\left(5x-1\right)^2=16\Leftrightarrow\left(5x-1\right)^2=4^2\) hoặc \(\left(5x-1\right)^2=\left(-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\left(l\right)\end{matrix}\right.\)

Chọn D. Số 7 là số nguyên tố.

Sửa các đáp án còn lại: 

A. Số 0 là số tự nhiên.

B. Số 1 là số tự nhiên.

C. Số 2 là số nguyên tố.