C-C-B-B-C

7,B

8,C

9,B

10,B

11,C

câu 12 bị lỗi r bn

7: B

8: C 

9: B 

10: B 

11: C 

12: lỗi 

d)

\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)

e)

\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)

f)

\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)

\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)

\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)

\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)

\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)

\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)

\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)

\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)

Đặt A = 1 + 3 + 5 + ... + 97 + 99

Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)

Tổng A bằng: (99 + 1) . 50 : 2 = 2500

Thay A = 2500 vào biểu thức (1), ta được:

\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)

a)

\(\dfrac{\left(-14\right).15}{21.\left(-10\right)}\\ =\dfrac{-7.2.3.5}{7.3.-2.5}\\=\dfrac{7.2.3.5}{7.2.3.5}\\ =1\)

b)

\(\dfrac{5.7-7.9}{7.2+6.7}\\ =\dfrac{7\left(5-9\right)}{7\left(2+6\right)}\\ =\dfrac{-4}{8}\\ =\dfrac{-2.2}{2.4}\\ =-\dfrac{1}{2}\)

c)

\(\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}\\ =\dfrac{-7.3+2.-7.2}{7\left(-5-2\right)}\\ =\dfrac{-7\left(3+4\right)}{7.-7}\\ =\dfrac{7}{7}\\ =1\)

A) \(\frac{7}{8}-\frac{5}{12}=\frac{84}{96}-\frac{40}{96}=\frac{44}{96}=\frac{11}{24}\)

B)\(\frac{8}{15}-\frac{2}{5}=\frac{8}{15}-\frac{6}{15}=\frac{2}{15}\)

C)\(\frac{5}{6}-\frac{3}{10}=\frac{50}{60}-\frac{18}{60}=\frac{32}{60}=\frac{8}{15}\)

D)\(\frac{7}{10}-\frac{1}{5}=\frac{7}{10}-\frac{2}{10}=\frac{5}{10}=\frac{1}{2}\)

2)A)\(\frac{5}{10}-\frac{2}{15}=\frac{1}{2}-\frac{2}{15}=\frac{15}{30}-\frac{4}{30}=\frac{11}{30}\)

B)\(\frac{5}{20}-\frac{1}{6}=\frac{1}{4}-\frac{1}{6}=\frac{6}{24}-\frac{4}{24}=\frac{2}{24}=\frac{1}{12}\)

C)\(\frac{6}{18}-\frac{6}{24}=\frac{1}{3}-\frac{1}{4}=\frac{4}{12}-\frac{3}{12}=\frac{1}{12}\)

D)\(\frac{5}{9}-\frac{3}{12}=\frac{5}{9}-\frac{1}{4}=\frac{20}{36}-\frac{9}{36}=\frac{11}{36}\)

Chúc bạn học tốt

a ) − 2 5 b ) 1 6 c ) − 3 4

d ) − 3 4 e ) − 1 f ) 1 2

nếu có số 100 ở đề bài thì có3 cặp

a) 

\(\dfrac{6}{12}=\dfrac{6:6}{12:6}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}=\dfrac{1\times2}{2\times2}=\dfrac{2}{4}\)

Mà \(\dfrac{2}{4}< \dfrac{3}{4}\)

Vậy \(\dfrac{6}{12}< \dfrac{3}{4}\).

b)

\(\dfrac{8}{10}=\dfrac{8:2}{10:2}=\dfrac{4}{5}\)

Mà \(\dfrac{2}{5}< \dfrac{4}{5}\)

Vậy \(\dfrac{2}{5}< \dfrac{8}{10}\).

c)

\(\dfrac{40}{35}=\dfrac{40:5}{35:5}=\dfrac{8}{7}\)

Mà \(\dfrac{8}{7}>\dfrac{6}{7}\)

Vậy \(\dfrac{40}{35}>\dfrac{6}{7}\).

d)

\(\dfrac{8}{16}=\dfrac{8:8}{16:8}=\dfrac{1}{2}\)

Mà \(\dfrac{1}{2}< \dfrac{5}{2}\)

Vậy \(\dfrac{8}{16}< \dfrac{5}{2}\).

Câu 3:

a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)

\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)

b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)

\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)

Câu 4: 

a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)

b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)

\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)

\(=-27\cdot100-23\cdot100\)

\(=100\left(-27-23\right)\)

\(=-50\cdot100=-5000\)

c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)

d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)

\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)

\(=-1+1+\dfrac{5}{6}\)

\(=\dfrac{5}{6}\)