4.D

5.C

6.C

7.C

\(-2^2+2.\left(-\dfrac{2}{3}\right)+5\dfrac{1}{3}\)

= \(-4+\dfrac{-4}{3}+5+\dfrac{1}{3}\)

= \(-4+\left(-1\right)+5\)

= 0

21. B

22. B

23. A

24. D

(m,n) = {(1,1)} (vi m, n la so nguyen duong)

a) 5, 1, 1/2, 0, -2/3, -2, -3

b) \(\sqrt{2}\), 1, 1/2, 0, -1, -4/3, \(-\sqrt{5}\), -3

\(\dfrac{-3}{4}\cdot\dfrac{12}{-5}\cdot\dfrac{-25}{6}\)

=\(\dfrac{-12}{4}\cdot\dfrac{-25}{5}\cdot\dfrac{-12}{6}\)

=\(-3\cdot\left(-5\right)\cdot\left(-2\right)\)

=\(-30\)

\(\dfrac{2021}{1\cdot5}+\dfrac{2021}{5\cdot9}+...+\dfrac{2021}{x\cdot\left(x+4\right)}=505\)

\(2021\cdot\left(\dfrac{1}{1.5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{x\cdot\left(x+4\right)}\right)=505\)

\(\dfrac{2021}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{x\cdot\left(x+4\right)}\right)=505\)

\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(1-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(\dfrac{1}{x+4}=\dfrac{1}{2021}\)

=> \(x+4=2021\)

=> \(x=2017\)

vậy \(x=2017\)

b) \(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)

=> \(3\left(x-2\right)=4\left(5+x\right)\)

\(3x-6=20+4x\)

\(3x-4x=20+6\)

\(-x=26\)

=> \(x=26\)

Vậy \(x=26\)

b) Vì \(2^2+3^2+...+12^2=649\)

\(=>3^2\left(2^2+3^2+...+12^2\right)=3^2\cdot649\)

\(=>6^2+9^2+...+36^2=5841\)

=> A = \(1+9+5841\) = 5851