E tk nha:

22. \(\dfrac{\dfrac{3}{4}+\dfrac{3}{7}-\dfrac{3}{8}}{\dfrac{5}{4}+\dfrac{5}{7}-\dfrac{5}{8}}+1=\dfrac{3\left(\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{1}{8}\right)}{5\left(\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{1}{8}\right)}+1=\dfrac{3}{5}+1=\dfrac{8}{5}\)

23. \(\dfrac{2}{5}+\dfrac{3}{5}:\left(\dfrac{3}{5}+\dfrac{-2}{3}\right)+\dfrac{7}{2}=\dfrac{2}{5}+\dfrac{3}{5}:\left(\dfrac{9-10}{15}\right)+\dfrac{7}{2}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{-15}{1}+\dfrac{7}{2}=\dfrac{2}{5}-\dfrac{45}{5}+\dfrac{7}{2}\)

\(=\dfrac{4}{10}-\dfrac{90}{10}+\dfrac{35}{10}=\dfrac{4-90+35}{10}=-\dfrac{51}{10}\)

24. \(\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}=\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5+\dfrac{5}{73}-\dfrac{5}{115}}=\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}=\dfrac{4}{5}\)

4. \(\dfrac{5}{7}-\dfrac{2}{11}+\dfrac{5}{7}-\dfrac{9}{11}=\dfrac{5}{7}\left(-\dfrac{2}{11}-\dfrac{9}{11}\right)=-\dfrac{5}{7}\)

9. \(\dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}.\dfrac{3}{13}=\dfrac{5}{9}\left(\dfrac{7}{13}+\dfrac{9}{13}-\dfrac{3}{13}\right)=\dfrac{5}{9}\)

10. \(\dfrac{-5}{9}+\dfrac{8}{15}-\dfrac{2}{11}-\dfrac{4}{9}+\dfrac{7}{15}=\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}=-\dfrac{2}{11}\)

bài đâu bạn

Câu 16: A

Câu 14: C

Câu 12: A

TL

12 : 1,25 : 8

= 12 : (1,25 x 8 )

= 12 : 10

=1,2

HT

Học tốt nha bạn^^

12 : 1,25 : 8

= 12 : (1,25 x 8 )

= 12 : 10

=1,2

tích cho tui nhé

=1,2 nhé

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)

\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)

\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)

\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)

\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)