Bài 1: 

Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)

Vậy: \(x=\dfrac{4}{65}\)

Bài 2: 

a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)

\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)

\(=\dfrac{5366}{527}\)

bạn lên mạng tra từng câu 1 sẽ có

ukm cảm ơn bạn nhìu

lớp 1 mà cậu

4.24.5²-(3³.18+3³.12)

=4.24.25-[27.(18+12)]

=(4.25).24-[27.30]

=100.24-810

=2400-810

=1590

4.24.5^2-(3^3.18+3^3.12)

31.15.7^2.4-31.49.40

1+2+3+4+5+6+7+8+9+10

1+3+5+7+9+11+13+15+19

2+6+10+14+22+23+26+34

5+8+11+14+17+20+23+26+29

1+6+11+16+21+26+31+36+41+47+51

10+13+16+19+22+25+28+31+34+37+40

5+7+9+11+13+15+17+3+8+13+18+23+28

4+7+10+13+16+19+5+9+13+17+21+25 = 1590

#HT#

a. 15.27-3.5.17

=15.27-15.17

=15.(27-17)

=15.10

=150

\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)

c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))

= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)

= (17 - 6)  - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))

= 11  - \(\dfrac{15}{17}\)+ 0

=    \(\dfrac{172}{17}\)

b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)

= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)

= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))

= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))

= 250 +  \(\dfrac{193}{140}\)

= 250\(\dfrac{193}{140}\)

a: =-3/24-40/24

=-43/24

b: \(=\dfrac{6}{54}\cdot\dfrac{49}{35}=\dfrac{1}{9}\cdot\dfrac{7}{5}=\dfrac{7}{45}\)

c: \(=\dfrac{6}{5}+\dfrac{4}{3}=\dfrac{18+20}{15}=\dfrac{38}{15}\)

d: \(=\dfrac{31}{17}-\dfrac{14}{17}-\dfrac{5}{13}-\dfrac{8}{13}=1-1=0\)