\(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4\cdot3^{12}+6^{11}}\) hay \(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4}\cdot3^{12}+6^{11}\)?

\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{1\cdot\left(3^2\right)^3-\left(3\cdot2\right)^9\cdot\left(5^2\cdot2^4\cdot3\right)}{\left(2^3\right)^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{3^6-3^9\cdot2^9\cdot5^2\cdot2^4\cdot3}{2^{12}\cdot3^{12}+\left(2\cdot3\right)^{11}}\)

= \(\dfrac{3^6-3^{10}\cdot2^{13}\cdot5^2}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

như lời đã nói thì bn làm tiếp>:)))

các bạn ơi các bạn có thể chỉ mình lời giải được ko thank các bạnh nhé

1 + 4 = 5

52 + 7 = 59

123 + 6 = 129 

218 + 11 = 229

1 + 4 = 5

52 + 7 = 59

123 + 6 = 129 

218 + 11 = 229

sai hết nhé

`c)(15/(sqrt6+1)+4/(sqrt6-2)-12/(3-sqrt6))*(sqrt6+11)`

`=((15(sqrt6-1))/(6-1)+(4(sqrt6+2))/(6-4)-(12(3+sqrt6))/(9-6))*(sqrt6+11)`

`=(3(sqrt6-1)+2(sqrt6+2)-4(3+sqrt6))*(sqrt6+11)`

`=(3sqrt6-3+2sqrt6+4-12-4sqrt6)*(sqrt6+11)`

`=(sqrt6-11)(sqrt6+11)`

`=6-121=-115`

c) Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121=-115

5/6=40/48

11/12=44/48

Ba phân số cần tìm là 41/48; 42/48; 43/48