giúp mình nhé

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)

\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)

\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)

Bạn ơi cho mk hỏi chỗ đoạn kia bạn  lấy 1-7 ở đâu và 1 + 8 ở đâu

\(B=\frac{2^{13}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{13}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)

\(=\frac{2^{13}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(=\frac{2^{12}.3^4\left(2.3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)

\(=\frac{2^{12}.3^4.5}{2^{12}.3^5.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)

\(=\frac{5}{12}-\frac{-10}{3}=\frac{5}{12}+\frac{40}{12}=\frac{45}{12}=\frac{15}{4}=3\frac{3}{4}\)

Câu a:

- |- 5 + 3 - 7| - |- 5 + 7|

= -|- 2 - 7| - |2|

= - |-9| - 2

= - 9 - 2

= -11

Câu b:

24 - (72 - 13 + 24) - (72 - 13)

= 24 - 72 + 13 - 24 - 72 + 13

= (24 - 24) - (72 + 72) + (13 + 13)

= 0 - 144 + 26

= - 118

Câu c:

|4 - 9 - 5| - (4 - 9 - 5) - 15 + 9

=|- 5 - 5| - (- 5 - 5) - 15 + 9

= |-10| - (-10) - 15 + 9

= 10 + 10 - 15 + 9

= 20 - 15+ 9

= 5 +9

= 14

Câu d:

- 20 - (25 - 11 + 8) +(25 - 8 + 20)

= - 20 - 25 + 11 - 8 + 25 - 8 + 20

= (-20 + 20) + (25 - 25) + (11 - 8- 8)

= 0 + 0+ 3 - 8

= -5

\(P=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6-2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot8}\)

\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3-1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(1-7\right)}{5^9\cdot7^3\cdot\left(1+8\right)}\)

\(=\dfrac{1}{3}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{3}{9}+\dfrac{30}{9}=\dfrac{33}{9}=\dfrac{11}{3}\)

\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)

\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\\ =\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\\ =\dfrac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\\ =\dfrac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\\ =\dfrac{-2}{3}-\dfrac{5.\left(-6\right)}{9}\\ =\dfrac{-2}{3}+\dfrac{30}{9}\\ =\dfrac{8}{3}\)