\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)

\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)

\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)

mk ko viết lại đề

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

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

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

eo ôi t làm rồi mà bị xoá :v thôi  t hướng dẫn :v

Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)

Ta có:

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)

\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.9^2}{12^6+8^4.3^5}+\frac{16^3.3^9.3+120.3^9.2^9}{4^6.\left(3^2\right)^6+6^{12}}\)

\(A=\frac{2^{12}.3^5-2^{12}.9^2}{3^6.4^6+8^4.3^5}+\frac{3^9.\left[16^3.3+120.2^9\right]}{4^6.3^{12}+3^{12}.2^{12}}\)

\(A=\frac{2^{12}.162}{3^5.3.4^6+3^5.8^4}+\frac{3^9.73728}{3^{12}.\left(4^6+2^{12}\right)}\)

\(A=\frac{2^{12}.162}{3^5.\left(3.4^6+8^4\right)}+\frac{3^9.73728}{3^{12}.8192}\)

         Đến đây bn tự làm nha hoa mắt lắm ròi

a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

Kotori Minami bạn có thể giải chi tiết ra duoc ko

o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)

s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)

\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)

\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)

t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)

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

\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)

\(B=\frac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}=\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}=\frac{2^{12}.\left(3^5-3^6\right)}{2^{12}.\left(3^6+3^5\right)}=\frac{3^5-3^6}{3^6+3^5}=\frac{3^5.\left(1-3\right)}{3^5.\left(3+1\right)}=\frac{1-3}{3+1}=-\frac{2}{4}=-\frac{1}{2}\)

\(\frac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}\)

\(=\frac{2^{12}.3^5-4^6.3.3^5}{2^{12}.3.3^5+4^6.3^5}\)

\(=\frac{1-3}{3+1}\)

\(=\frac{-1}{2}\)

P=2¹².3⁵-4⁶.3⁶/2¹².9³+8⁴.3⁵

P=2¹².3⁵-(2²)⁶.3⁶/2¹².(3³)³+(2³)⁴.3⁵

P=2¹².3⁵-2¹².3⁶/2¹².3⁹+2¹².3⁵

P=2¹².(3⁵-3⁶)/2¹².(3⁹+3⁵)

P=3⁵-3⁶/3⁹+3⁵

= 2¹⁰⁰ - (2⁹⁹ + 2⁹⁸ + ... + 2² + 2)

Đặt A = 2⁹⁹ + 2⁹⁸ + ... + 2² + 2

2A = 2¹⁰⁰ + 2⁹⁹ + ... + 2³ + 2²

2A - A = (2¹⁰⁰ + 2⁹⁹ + ... + 2³ + 2²) - (2⁹⁹ + 2⁹⁸ + ... + 2² + 2)

A = 2¹⁰⁰ - 2

Ta có:

2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2

= 2¹⁰⁰ - (2¹⁰⁰ - 2)

= 2¹⁰⁰ - 2¹⁰⁰ + 2

= 0 + 2

= 2

= 3² + 2⁸ : 2⁹

\(=9+\frac{1}{2}\)

\(=\frac{18}{2}+\frac{1}{2}=\frac{19}{2}\)