\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

Phân tích từ số:

\(\frac{2^{12}.3^5-4^2.4^4.3^4}{2^{12}.3.3^5+4^2.4^4.3.3^4}=\frac{1}{6}\)

\(\frac{5^9.5.7^3-5^9.5.7^3.7}{5^9.7^3+5^9.2^3.7^3}=\frac{-10}{3}\)

Sau khi rút gọn là:

\(\frac{1}{6}-\left(-\frac{10}{3}\right)=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)

a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)

\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)

\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)

\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)

b,\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)

\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)

\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)

\(=3^n\cdot9+1-2^n\cdot4+1\)

\(=3^n\cdot10-2^n\cdot5\)

Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)

\(3^n\cdot10⋮10\)

Vậy : ....

A = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

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

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

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

  = \(\frac{1}{6}-\frac{-10}{3}\)

  = 7/2

bố mầy đéo hiểu cái éo gì

12fgergtefe

Gợi ý :  Phân tích hết ra thành tích các thừa số nguyên tố rồi đặt cái chung ra ngoài

-> rút gọn

-> kết quả

P/S : bài này cx ko dài lắm nhưg lười  ^^

ko được nói tục

Ta có \(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.7^3.2^3}\)

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

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

\(=\frac{2}{12}-\frac{5\left(-6\right)}{9}\)\(=\frac{1}{6}+\frac{30}{9}=\frac{7}{2}\)

Vậy A=7/2

\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

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

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

\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)

\(=\dfrac{1}{6}-\dfrac{-10}{3}\)

\(=\dfrac{7}{2}\)

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

    = \(\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\) = \(\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}\)= \(\frac{1}{6}-\frac{-10}{3}\)= 1/6 + 10/3 = 7/2

7/2 la dung

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

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)

=2^12.3^4.(3-1)/2^12.3^5(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+2^3)

2/3.4-5.(-6)/9

=1/6-(-10/3)

1/6+10/3

7/2

What ???????????????