a ,b

cách 1 : cộng 2 số trong ngặc rồi tính bình thường 

cách 2 : nhân phân phối ra rồi tính nhân chia trước cộng trừ sau

c,d

cách 1 : tính bình thường, nhân chia trc cộng trừ sau

cách 2 : đặt nhân tử ( thừa số ) chung

Cách 1 : \((\frac{6}{11}+\frac{5}{11})\cdot\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}\)

Cách 2 : \(\frac{6}{11}\cdot\frac{3}{7}+\frac{5}{11}\cdot\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)

Cách 1 dễ rồi nha

Cách 2 : \(\frac{6}{7}:\frac{2}{5}-\frac{4}{7}:\frac{2}{5}=\frac{15}{7}-\frac{10}{7}=\frac{5}{7}\)

Cách 1 tự làm

Cách 2 : \(\frac{3}{5}\cdot(\frac{7}{9}-\frac{2}{9})=\frac{3}{5}\cdot\frac{5}{9}=\frac{1}{3}\)

Cách 1 : ...

Cách 2 : \((\frac{8}{15}+\frac{7}{15}):\frac{2}{11}=1:\frac{2}{11}=\frac{2}{11\cdot1}=\frac{2}{11}\)

`a) 5 / 13 xx 4 / 15 xx 13 = [ 5 xx 4 xx 13 ] / [ 13 xx 5 xx 3 ] = 4 / 3`

`b) ( 3 / 7 + 5 / 2 ) xx 7 / 5 = 3 / 7 xx 7 / 5 + 5 / 2 xx 7 / 5 = 3 / 5 + 7 / 2 = 6 / 10 + 35 / 10 = 41 / 10`

`c) 1 / 5 xx 11 / 18 + 11 / 18 xx 3 / 5 = 11 / 18 xx ( 1 / 5 + 3 / 5 ) = 11 / 18 xx 4 / 5 = [ 11 xx 2 xx 2 ] / [ 2 xx 9 xx 5 ] = 22 / 45`

\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{4}{3}\)

\(b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{6}{14}+\dfrac{35}{14}\right)\times\dfrac{7}{5}=\dfrac{41}{14}\times\dfrac{7}{15}=\dfrac{41}{30}\)

\(c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)

a) 11^5

b) 17^2 

c) 4^2 

d) a^4 

b: \(27D=3^{14}+3^{17}+...+3^{2024}\)

\(\Leftrightarrow26D=3^{2024}-3^{11}\)

hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)

c: \(25E=-5^4-5^6-...-5^{1002}\)

\(\Leftrightarrow24E=-5^{1002}+5^2\)

hay \(E=\dfrac{-5^{1002}+5^2}{24}\)

sai nha

a: \(=\dfrac{9}{4}+\dfrac{11}{14}=\dfrac{63+22}{28}=\dfrac{85}{28}\)

b: \(=\dfrac{-7}{5}-\dfrac{11}{5}=-\dfrac{18}{5}\)

c: \(=\dfrac{-12}{13}-\dfrac{9}{13}=\dfrac{-21}{13}\)

d: \(=\dfrac{3-5}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}\)

a: \(\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-3^{10}\cdot5^{21}}{3^{12}\cdot5^{20}}=\frac{-5}{3^2}=-\frac59\)

b: \(\frac{-11^5\cdot13^7}{11^5\cdot13^8}=\frac{-11^5}{11^5}\cdot\frac{13^7}{13^8}=-\frac{1}{13}\)

c: \(2^{10}\cdot3^{10}-2^{10}\cdot3^9\)

\(=2^{10}\cdot3^9\cdot\left(3-1\right)\)

\(=2^{11}\cdot3^9\)

Ta có: \(\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)

\(=\frac{2^{11}\cdot3^9}{2^{10}\cdot3^9}=2\)

d: \(5^{11}\cdot7^{12}+5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(7+1\right)=5^{11}\cdot7^{11}\cdot8\)

\(5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(5\cdot7+9\right)\)

\(=5^{11}\cdot7^{11}\left(35+9\right)=5^{11}\cdot7^{11}\cdot44\)

Ta có: \(\frac{5^{11}\cdot7^{12}+5^{11}\cdot7^{11}}{5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}}\)

\(=\frac{5^{11}\cdot7^{11}\cdot8}{5^{11}\cdot7^{11}\cdot44}=\frac{8}{44}=\frac{2}{11}\)

a: =(5/7+2/7)+(4/3+5/3)=3+1=4

b: =(17/12+7/12)+(29/7-8/7)

=2+3=5

c: =(2/5+3/5)+(6/9+1/3)+(7/4+1/4)

=1+2+1

=4

d: =(1/3+2/3)+(13/17+4/17)+(29/11+4/11)

=1+1+3=5