37 và 73

0,2 và \(\frac{3}{10}\)

Ta có: \(\frac{3}{10}\) = 0,3

=> Ta so sánh 0,2 và 0,3

Do đó 0,2 > 0,3.

ta có : \(\frac{3}{10}=0,3>0,2\)
nên \(\frac{3}{10}>0,2\)

0,3579 > 0,33579

0,3579 > 0,33579

`x - 5,01 = 7,02 - 2 xx 1,5`

`=>x - 5,01 = 7,02 - 3`

`=> x - 5,01 = 4,02`

`=> x = 4,02 + 5,01`

`=> x  = 9,03`

Vậy `x = 9,03`

X - 5,01=4,02

X=4,02+5,01

X=9,03

\(x-\dfrac{3}{4}=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{4}+\dfrac{3}{5}=\dfrac{15}{20}+\dfrac{12}{20}=\dfrac{27}{20}\)

\(x\) - \(\frac34\) = \(\frac35\)

\(x=\frac35\) + \(\frac34\)

\(x=\) \(\frac{12}{20}\) + \(\frac{15}{20}\)

\(x\) = \(\frac{27}{20}\)

Vậy \(x=\frac{27}{20}\)

\(\frac{3}{-8}\) < 0 < \(\frac58\)

Vậy \(\frac{3}{-8}\) < \(\frac58\)

\(\dfrac{3}{-8}< 0;0< \dfrac{5}{8}\)

Do đó: \(\dfrac{3}{-8}< \dfrac{5}{8}\)

Vì 2022<2023<2024<2025

nên \(\dfrac{1}{2022}>\dfrac{1}{2023}>\dfrac{1}{2024}>\dfrac{1}{2025}\)

=>PHân số nhỏ nhất là \(\dfrac{1}{2025}\)

\(\dfrac{3}{2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{2024\cdot2025}\\ =3\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2024}-\dfrac{1}{2025}\right)\\ =3\cdot\left(\dfrac{1}{1}-\dfrac{1}{2025}\right)\\ =3\cdot\dfrac{2024}{2025}=\dfrac{2024}{675}\)

= \(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{2024.2025}\right)\)
= 3.(\(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\cdots+\frac{2025-2024}{2024.2025}\) )
=3.\(\left(\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2024}-\frac{1}{2025}\right)\)
=3.\(\left(1-\frac{1}{2025}\right)\)
=3.\(\frac{2024}{2025}\)
=\(\frac{2024}{675}\)

2 - (2\(x\) + \(\frac14\)) \(\times\) \(\frac{1}{10}\) = 20

(2\(x+\frac14\)) \(\times\) \(\frac{1}{10}\) = 2 - 20

(2\(x\) + \(\frac14\)) \(\times\frac{1}{10}\) = -18

(2\(x+\frac14\)) = - 18 x 10

2\(x+\frac14=-180\)

2\(x\) = - 180 - \(\frac14\)

2\(x\) = - \(\frac{721}{4}\)

\(x=-\frac{721}{4}:2\)

\(x=-\frac{721}{8}\)

Vậy \(x=-\frac{721}{8}\)

Bằng 1